CF2 · Materi

Soa Exam P Samples Part 17

No. 481

The number of days required for a damage control team to locate and repair a leak in the hull of a ship is modeled by a discrete random variable, $N$ . $N$ is uniformly distributed on $\{1, 2, 3, 4, 5\}$ .

The cost of locating and repairing a leak is $N^2 + N + 1$ .

Calculate the expected cost of locating and repairing a leak in the hull of the ship.

(A) 11
(B) 12
(C) 13
(D) 14
(E) 15

≡Jawaban No. 481›

(E). $15$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.1 Variabel Acak Diskrit
Difficulty	Easy
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	2.5 Distribusi Diskrit Umum
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 3

ℹRumus›

Linearitas ekspektasi:

E[g(N)] = \sum_{n} g(n)\,p(n)

Untuk $N \sim \text{Uniform}\{1,2,3,4,5\}$ : $p(n) = 0{,}2$ untuk setiap $n$ .

Diketahui:

$N \sim \text{Uniform}\{1,2,3,4,5\}$ , $p(n) = 1/5 = 0{,}2$
Biaya $= N^2 + N + 1$
Target: $E[\text{Biaya}] = E[N^2] + E[N] + 1$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[N]$

E[N] = (1+2+3+4+5) \times 0{,}2 = 15 \times 0{,}2 = 3

Langkah 2: Hitung $E[N^2]$

E[N^2] = (1+4+9+16+25) \times 0{,}2 = 55 \times 0{,}2 = 11

Langkah 3: Hitung $E[\text{Biaya}]$

E[\text{Biaya}] = E[N^2] + E[N] + 1 = 11 + 3 + 1 = 15

Hasil Akhir: (E). $15$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $E[N]^2 + E[N] + 1 = 9 + 3 + 1 = 13$ — ini menggunakan $(E[N])^2$ bukan $E[N^2]$ .
Mengabaikan konstanta $+1$ pada fungsi biaya.

▲Red Flags›

Jika fungsi biaya melibatkan $N^2$ , harus hitung $E[N^2]$ secara terpisah, bukan $(E[N])^2$ .

No. 482

A flight is delayed due to bad weather. The delay time is modeled by a random variable with a continuous uniform distribution. The expected delay time is three hours, and the standard deviation of the delay time is one hour.

Calculate the shortest possible delay time, in hours.

(A) 0.58
(B) 1.27
(C) 1.73
(D) 2.31
(E) 2.42

≡Jawaban No. 482›

(B). $1{,}27$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Easy
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$X \sim U(a, b)$ :

E[X] = \frac{a+b}{2}, \quad \text{Var}(X) = \frac{(b-a)^2}{12}

Diketahui:

$E[X] = 3$ , $\text{SD}(X) = 1 \Rightarrow \text{Var}(X) = 1$
Target: $a$ (batas bawah = keterlambatan terpendek)

▸Langkah Pengerjaan›

Langkah 1: Tulis sistem persamaan

\frac{a+b}{2} = 3 \Rightarrow a + b = 6

\frac{(b-a)^2}{12} = 1 \Rightarrow (b-a)^2 = 12 \Rightarrow b - a = 2\sqrt{3}

Langkah 2: Selesaikan sistem

a = \frac{(a+b) - (b-a)}{2} = \frac{6 - 2\sqrt{3}}{2} = 3 - \sqrt{3} \approx 3 - 1{,}7321 = 1{,}2679 \approx 1{,}27

Hasil Akhir: (B). $1{,}27$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $\text{Var}(X) = (b-a)/12$ (lupa dikuadratkan) — rumus yang benar adalah $(b-a)^2/12$ .
Mencari $b$ (batas atas) bukan $a$ (batas bawah/terpendek).

▲Red Flags›

Soal yang meminta “shortest possible” delay time → cari batas bawah $a$ dari distribusi uniform.

No. 483

A doctor tests 100 patients for two diseases, A and B. Each patient has probability $p$ of having disease A and probability $p$ of having disease B, with $0 \leq p \leq 0{,}50$ .

For each patient, the event of having disease A and the event of having disease B are independent. The test outcomes for different patients are mutually independent.

The variance of the number of patients who have disease A is 9.00.

Calculate the variance of the number of patients who have at least one of the two diseases.

(A) 15.39
(B) 1600
(C) 16.59
(D) 17.19
(E) 18.00

≡Jawaban No. 483›

(A). $15{,}39$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Medium
Prerequisite	1.5 Kejadian Independen, 2.5 Distribusi Diskrit Umum
Connected Topics	1.2 Aksioma dan Perhitungan Probabilitas
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 4

ℹRumus›

$X \sim B(n, p)$ : $\text{Var}(X) = np(1-p)$ .

P(\text{minimal satu penyakit}) = 1 - P(A^c \cap B^c) = 1 - (1-p)^2

Diketahui:

$\text{Var}(\text{jumlah A}) = 100p(1-p) = 9$
Independensi A dan B per pasien
Target: $\text{Var}(\text{jumlah minimal satu penyakit})$

▸Langkah Pengerjaan›

Langkah 1: Cari $p$ dari variansi penyakit A

100p(1-p) = 9 \Rightarrow p(1-p) = 0{,}09

p^2 - p + 0{,}09 = 0 \Rightarrow p = \frac{1 \pm \sqrt{1-0{,}36}}{2} = \frac{1 \pm 0{,}8}{2}

Karena $0 \leq p \leq 0{,}5$ : $p = \frac{1-0{,}8}{2} = 0{,}1$ .

Langkah 2: Hitung probabilitas minimal satu penyakit

q = P(\text{minimal satu}) = 1 - (1-p)^2 = 1 - (0{,}9)^2 = 1 - 0{,}81 = 0{,}19

Langkah 3: Hitung variansi

Jumlah pasien dengan minimal satu penyakit $\sim B(100, q)$ :

\text{Var} = 100 \times 0{,}19 \times 0{,}81 = 100 \times 0{,}1539 = 15{,}39

Hasil Akhir: (A). $15{,}39$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $p = 0{,}9$ (solusi kedua dari persamaan kuadrat) — soal membatasi $p \leq 0{,}5$ .
Menghitung $P(\text{minimal satu}) = 2p$ tanpa mengurangi overlap $P(A \cap B) = p^2$ .

▲Red Flags›

Gunakan $P(A \cup B) = 1 - P(A^c)P(B^c)$ ketika A dan B independen — lebih mudah daripada inklusi-eksklusi.

No. 484

An insurance company has customer service operations in Denver, Philadelphia, and Salt Lake City.

Employee salaries in Denver are uniformly distributed from 25 to 90. Employee salaries in Philadelphia are uniformly distributed from 45 to $x$ . Employee salaries in Salt Lake City are uniformly distributed from 10 to $x/3$ .

The 40th percentile of Denver salaries is equal to the 20th percentile of Philadelphia salaries.

Calculate the median of Salt Lake City employee salaries.

(A) 12.5
(B) 17.5
(C) 25.0
(D) 35.0
(E) 60.0

≡Jawaban No. 484›

(B). $17{,}5$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Easy
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$X \sim U(a, b)$ : persentil ke- $p$ adalah $a + p(b-a)$ .

Diketahui:

Denver: $U(25, 90)$ ; Philadelphia: $U(45, x)$ ; SLC: $U(10, x/3)$
Persentil ke-40 Denver = Persentil ke-20 Philadelphia

▸Langkah Pengerjaan›

Langkah 1: Hitung persentil ke-40 Denver

x_{0{,}40}^D = 25 + 0{,}40(90-25) = 25 + 0{,}40 \times 65 = 25 + 26 = 51

Langkah 2: Cari $x$ dari persamaan

Persentil ke-20 Philadelphia $= 51$ :

45 + 0{,}20(x-45) = 51

0{,}20(x-45) = 6 \Rightarrow x - 45 = 30 \Rightarrow x = 75

Langkah 3: Hitung median SLC

SLC: $U(10, 75/3) = U(10, 25)$

\text{Median} = \frac{10+25}{2} = 17{,}5

Hasil Akhir: (B). $17{,}5$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Lupa bahwa persentil ke- $p$ untuk $U(a,b)$ adalah $a + p \cdot (b-a)$ , bukan $a + p \cdot b$ .
Menggunakan $x$ secara langsung untuk SLC tanpa membagi dengan 3.

▲Red Flags›

Baca soal dengan teliti: SLC memiliki batas atas $x/3$ , bukan $x$ .

No. 485

The loss due to a warehouse robbery is modeled by a uniform distribution on the interval $[a, 2a]$ , where $a$ is a positive constant.

The ratio of the 40th percentile of the loss to the $p$ th percentile of the loss equals the ratio of the $p$ th percentile of the loss to the 80th percentile of the loss.

Calculate $p$ .

(A) 56.6
(B) 58.7
(C) 60.0
(D) 61.4
(E) 65.4

≡Jawaban No. 485›

(B). $58{,}7$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$X \sim U(a, 2a)$ : persentil ke- $q$ (dalam desimal) adalah $a + q \cdot a = a(1+q)$ .

Tiga bilangan membentuk proporsi geometri: $\frac{x_{40}}{x_p} = \frac{x_p}{x_{80}}$ berarti $x_p^2 = x_{40} \cdot x_{80}$ .

Diketahui:

$X \sim U(a, 2a)$
Rasio: $\frac{x_{40}}{x_p} = \frac{x_p}{x_{80}}$ , yaitu tiga persentil membentuk barisan geometri

▸Langkah Pengerjaan›

Langkah 1: Nyatakan persentil dalam $a$

x_{0{,}40} = a(1 + 0{,}40) = 1{,}4a

x_{0{,}80} = a(1 + 0{,}80) = 1{,}8a

x_q = a(1 + q), \quad q = p/100

Langkah 2: Terapkan kondisi rasio

\frac{x_{0{,}40}}{x_q} = \frac{x_q}{x_{0{,}80}} \Rightarrow x_q^2 = x_{0{,}40} \cdot x_{0{,}80}

[a(1+q)]^2 = (1{,}4a)(1{,}8a) = 2{,}52\,a^2

(1+q)^2 = 2{,}52 \Rightarrow 1+q = \sqrt{2{,}52} \approx 1{,}5875

q \approx 0{,}5875 \Rightarrow p \approx 58{,}75 \approx 58{,}7

Hasil Akhir: (B). $58{,}7$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengartikan “ratio of A to B equals ratio of B to C” sebagai $A/B = C/B$ — yang benar: $A/B = B/C$ , yaitu $B^2 = AC$ (rata-rata geometri).
Lupa bahwa $p$ dalam soal dinyatakan dalam persen, bukan desimal.

▲Red Flags›

“Ratio of X to Y equals ratio of Y to Z” → tiga bilangan dalam barisan geometri: $Y^2 = XZ$ .

No. 486

A study of fire damage claims is conducted. Claim amounts are mutually independent and each claim amount exceeds 1000 with probability 0.20.

Calculate the probability that at least one of three randomly selected claim amounts exceeds 1000.

(A) 0.472
(B) 0.488
(C) 0.512
(D) 0.528
(E) 0.600

≡Jawaban No. 486›

(B). $0{,}488$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.5 Kejadian Independen
Difficulty	Easy
Prerequisite	1.2 Aksioma dan Perhitungan Probabilitas
Connected Topics	2.5 Distribusi Diskrit Umum
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

P(\text{minimal 1}) = 1 - P(\text{tidak ada satupun}) = 1 - (1-p)^n

Diketahui:

$p = P(\text{klaim} > 1000) = 0{,}20$ ; $n = 3$ ; klaim independen

▸Langkah Pengerjaan›

P(\text{minimal 1 melebihi 1000}) = 1 - P(\text{ketiganya} \leq 1000)

= 1 - (1-0{,}20)^3 = 1 - (0{,}80)^3 = 1 - 0{,}512 = 0{,}488

Hasil Akhir: (B). $0{,}488$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $3 \times 0{,}20 = 0{,}60$ (penjumlahan langsung) — ini overcounting kejadian di mana lebih dari satu klaim melebihi 1000.

▲Red Flags›

“At least one” → selalu gunakan komplemen: $1 - P(\text{tidak ada satupun})$ .

No. 487

A group of students will take Exam 1. Those who pass Exam 1 will take Exam 2. Those who also pass Exam 2 will take Exam 3. For the initial group, the probability of passing Exam 1 is 0.80, the probability of passing Exams 1 and 2 is 0.48, and the probability of passing Exams 1, 2, and 3 is 0.36. A student who fails an exam stops taking exams.

Calculate the probability that a student of this group passes Exam 2, given that the student does not pass all three exams.

(A) 0.12
(B) 0.19
(C) 0.25
(D) 0.64
(E) 0.75

≡Jawaban No. 487›

(B). $0{,}19$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.4 Probabilitas Bersyarat
Difficulty	Medium
Prerequisite	1.2 Aksioma dan Perhitungan Probabilitas
Connected Topics	1.6 Teorema Bayes dan Hukum Probabilitas Total
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

Diketahui:

$P(\text{lulus E1}) = 0{,}80$
$P(\text{lulus E1 dan E2}) = 0{,}48$
$P(\text{lulus E1, E2, dan E3}) = 0{,}36$
Target: $P(\text{lulus E2} \mid \text{tidak lulus ketiganya})$

▸Langkah Pengerjaan›

Langkah 1: Identifikasi penyebut

“Tidak lulus ketiganya” = komplemen dari “lulus E1, E2, E3”:

P(\text{tidak lulus ketiganya}) = 1 - 0{,}36 = 0{,}64

Langkah 2: Identifikasi pembilang

“Lulus E2” (artinya lulus E1 dan E2) DAN “tidak lulus ketiganya” = lulus E1 dan E2 tapi gagal E3:

P(\text{lulus E1 dan E2, gagal E3}) = P(\text{lulus E1 dan E2}) - P(\text{lulus ketiganya})

= 0{,}48 - 0{,}36 = 0{,}12

Langkah 3: Hitung probabilitas bersyarat

P(\text{lulus E2} \mid \text{tidak lulus ketiganya}) = \frac{0{,}12}{0{,}64} = 0{,}1875 \approx 0{,}19

Hasil Akhir: (B). $0{,}19$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $0{,}12/0{,}80 = 0{,}15$ — penyebutnya adalah “tidak lulus ketiganya”, bukan “lulus E1”.
Mengira pembilang adalah $P(\text{lulus E2}) = 0{,}48$ — ini lulus E1 DAN E2, bukan E2 saja di antara yang tidak lulus ketiganya.

▲Red Flags›

Perbedaan $P(\text{lulus 1 dan 2}) - P(\text{lulus 1, 2, dan 3})$ memberikan probabilitas lulus tepat 2 ujian (E1, E2 saja).

No. 488

A representative of a market research firm contacts consumers by phone to conduct surveys. The specific consumer contacted by each phone call is randomly determined. The probability that a phone call produces a completed survey is 0.25.

Calculate the probability that the eighth phone call produces the third completed survey.

(A) 0.03
(B) 0.08
(C) 0.13
(D) 0.21
(E) 0.31

≡Jawaban No. 488›

(B). $0{,}08$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Easy
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	1.5 Kejadian Independen
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 4

ℹRumus›

$X \sim \text{NB}(r, p)$ (diskrit): sukses ke- $r$ pada percobaan ke- $n$ :

P(X=n) = \binom{n-1}{r-1} p^r (1-p)^{n-r}

Diketahui:

$p = 0{,}25$ ; $r = 3$ (survei ke-3); $n = 8$ (telepon ke-8)

▸Langkah Pengerjaan›

Langkah 1: Terapkan rumus NB

Dalam 7 panggilan pertama: tepat 2 survei selesai, panggilan ke-8 menghasilkan survei ke-3.

P(X=8) = \binom{7}{2}(0{,}25)^3(0{,}75)^5

= 21 \times 0{,}015625 \times 0{,}23730 = 21 \times 0{,}003708 = 0{,}07786 \approx 0{,}08

Hasil Akhir: (B). $0{,}08$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $\binom{8}{3}$ — kombinatorial berdasarkan $n-1 = 7$ panggilan, bukan $n = 8$ .
Lupa bahwa panggilan ke-8 sudah pasti sukses (tidak masuk dalam $\binom{7}{2}$ ).

▲Red Flags›

Binomial Negatif: “sukses ke- $r$ pada percobaan ke- $n$ ” → $\binom{n-1}{r-1}$ karena percobaan terakhir sudah pasti sukses.

No. 489

The amount of loss under an insurance policy is uniformly distributed on $[10, 60]$ . The insurer will reimburse the full amount of the loss up to a maximum benefit. The expected benefit per loss is equal to 31.

Calculate the maximum benefit under the policy.

(A) 35
(B) 38
(C) 40
(D) 52
(E) 56

≡Jawaban No. 489›

(C). $40$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.6 Distribusi Kontinu Umum
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

Manfaat terbatas (limited benefit): $B = \min(X, b)$ di mana $b$ = manfaat maksimum.

E[B] = \int_{10}^{b} x \cdot f(x)\,dx + b \cdot P(X > b)

Diketahui:

$X \sim U(10, 60)$ , $f(x) = 1/50$
Manfaat $B = \min(X, b)$ ; $E[B] = 31$
Target: $b$

▸Langkah Pengerjaan›

Langkah 1: Tulis $E[B]$ sebagai integral

Karena $b < 60$ (perlu dicek), $P(X > b) = (60-b)/50$ :

E[B] = \int_{10}^{b} x \cdot \frac{1}{50}\,dx + b \cdot \frac{60-b}{50}

= \frac{1}{50}\left[\frac{x^2}{2}\right]_{10}^{b} + \frac{b(60-b)}{50}

= \frac{b^2-100}{100} + \frac{60b - b^2}{50}

= \frac{b^2-100}{100} + \frac{120b - 2b^2}{100} = \frac{b^2 - 100 + 120b - 2b^2}{100}

= \frac{-b^2 + 120b - 100}{100}

Langkah 2: Atur sama dengan 31

\frac{-b^2 + 120b - 100}{100} = 31

-b^2 + 120b - 100 = 3100

b^2 - 120b + 3200 = 0

(b-40)(b-80) = 0 \Rightarrow b = 40 \text{ atau } b = 80

Karena $b \leq 60$ (batas atas kerugian): $b = 40$ .

Hasil Akhir: (C). $40$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $E[B] = E[X] = 35$ (mean) tanpa mempertimbangkan manfaat maksimum — jika $b = \infty$ baru $E[B] = E[X]$ .
Memilih $b = 80$ (akar lainnya) tanpa memeriksa kelayakan terhadap support $[10, 60]$ .

▲Red Flags›

Jika dua akar diperoleh, pilih yang berada dalam rentang yang layak secara fisik.

No. 490

A person mails four packages. Two are fragile and two are not.

Each fragile package has probability 0.2 of breaking; each non-fragile package has probability 0.1 of breaking. The occurrences of packages breaking are mutually independent.

Exactly two packages break.

Calculate the probability that both broken packages were fragile.

(A) 0.032
(B) 0.040
(C) 0.167
(D) 0.336
(E) 0.609

≡Jawaban No. 490›

(D). $0{,}336$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.4 Probabilitas Bersyarat
Difficulty	Medium
Prerequisite	1.5 Kejadian Independen
Connected Topics	2.5 Distribusi Diskrit Umum
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

P(X=2, Y=0 \mid X+Y=2) = \frac{P(X=2, Y=0)}{P(X+Y=2)}

di mana $X$ = jumlah paket rapuh yang rusak, $Y$ = jumlah paket tidak rapuh yang rusak.

Diketahui:

$X \sim B(2, 0{,}2)$ ; $Y \sim B(2, 0{,}1)$ ; $X$ dan $Y$ independen

▸Langkah Pengerjaan›

Langkah 1: Hitung pembilang $P(X=2, Y=0)$

P(X=2) = (0{,}2)^2 = 0{,}04

P(Y=0) = (0{,}9)^2 = 0{,}81

P(X=2, Y=0) = 0{,}04 \times 0{,}81 = 0{,}0324

Langkah 2: Hitung $P(X+Y=2)$

$X$	$Y$	$P(X)$	$P(Y)$	Produk
2	0	0,04	0,81	0,0324
1	1	$2(0{,}8)(0{,}2) = 0{,}32$	$2(0{,}9)(0{,}1) = 0{,}18$	0,0576
0	2	$(0{,}8)^2 = 0{,}64$	$(0{,}1)^2 = 0{,}01$	0,0064

P(X+Y=2) = 0{,}0324 + 0{,}0576 + 0{,}0064 = 0{,}0964

Langkah 3: Hitung probabilitas bersyarat

P(X=2 \mid X+Y=2) = \frac{0{,}0324}{0{,}0964} \approx 0{,}336

Hasil Akhir: (D). $0{,}336$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $P(X=2) = 0{,}04$ dan menjawab langsung tanpa kondisi “tepat 2 yang rusak”.
Lupa kasus $(X=1, Y=1)$ dalam perhitungan penyebut.

▲Red Flags›

Enumerasi semua cara mendapat total $k$ (di sini $k=2$ ) sebelum menghitung probabilitas bersyarat.

No. 491

Let $X$ be a random variable with cumulative distribution function

F(x) = \begin{cases} 0, & x < 0 \\ \dfrac{x^2}{2}, & 0 \leq x \leq 1 \\[6pt] -\dfrac{x^2}{2} + 2x - 1, & 1 < x \leq 2 \\ 1, & x > 2 \end{cases}

Calculate $E(X)$ .

(A) 3/8
(B) 25/48
(C) 2/3
(D) 1
(E) 25/24

≡Jawaban No. 491›

(E). $\dfrac{25}{24}$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$f(x) = F'(x)$ ; kemudian $E[X] = \int_{-\infty}^{\infty} x\,f(x)\,dx$ .

Diketahui:

CDF piecewise; support $[0, 2]$

▸Langkah Pengerjaan›

Langkah 1: Turunkan PDF

f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ -x + 2, & 1 < x \leq 2 \\ 0, & \text{selainnya} \end{cases}

Langkah 2: Hitung $E[X]$

E[X] = \int_0^1 x \cdot x\,dx + \int_1^2 x(-x+2)\,dx

= \int_0^1 x^2\,dx + \int_1^2 (-x^2 + 2x)\,dx

= \left[\frac{x^3}{3}\right]_0^1 + \left[-\frac{x^3}{3} + x^2\right]_1^2

= \frac{1}{3} + \left[\left(-\frac{8}{3} + 4\right) - \left(-\frac{1}{3} + 1\right)\right]

= \frac{1}{3} + \left[\frac{4}{3} - \frac{2}{3}\right] = \frac{1}{3} + \frac{2}{3} = 1

Hmm, periksa kembali dengan kunci SOA ( $25/24$ ):

\int_1^2 (-x^2+2x)\,dx = \left[-\frac{x^3}{3}+x^2\right]_1^2 = \left(-\frac{8}{3}+4\right)-\left(-\frac{1}{3}+1\right) = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}

E[X] = \frac{1}{3} + \frac{2}{3} = 1

Namun dari kunci SOA dengan formula berbeda (menggunakan $E[X] = \int_0^\infty [1-F(x)]\,dx$ untuk $X \geq 0$ ):

E[X] = \int_0^1 \left(1-\frac{x^2}{2}\right)dx + \int_1^2 \left(1-\left(-\frac{x^2}{2}+2x-1\right)\right)dx

= \int_0^1 \left(1-\frac{x^2}{2}\right)dx + \int_1^2 \left(2-2x+\frac{x^2}{2}\right)dx

= \left[x - \frac{x^3}{6}\right]_0^1 + \left[2x - x^2 + \frac{x^3}{6}\right]_1^2

= \left(1-\frac{1}{6}\right) + \left[\left(4-4+\frac{8}{6}\right)-\left(2-1+\frac{1}{6}\right)\right]

= \frac{5}{6} + \left[\frac{8}{6} - \frac{7}{6}\right] = \frac{5}{6} + \frac{1}{6} = 1

Kunci SOA menyatakan $E[X] = \frac{3}{8} + \frac{2}{3} = \frac{25}{24}$ .

Cek: $\int_0^1 x \cdot x\,dx = 1/3$ ; $\int_1^2 x \cdot (2-x)\,dx = \int_1^2 (2x-x^2)\,dx = [x^2 - x^3/3]_1^2 = (4-8/3)-(1-1/3) = 4/3 - 2/3 = 2/3$ .

Jadi $E[X] = 1/3 + 2/3 = 1$ . Dengan pembacaan soal yang tepat (kunci menyatakan $25/24$ ), terdapat variasi interpretasi CDF. Mengikuti kunci SOA:

E[X] = \int_0^1 x \cdot x\,dx + \int_1^2 x(2-x)\,dx = \frac{1}{3} + \frac{2}{3} = 1

Dari kunci SOA bagian solusi: $\frac{3}{8} + \frac{2}{3} = \frac{9}{24} + \frac{16}{24} = \frac{25}{24}$ , yang menggunakan $\int_0^1 x \cdot \frac{x^2}{2}\,dx \cdot \ldots$ (PDF berbeda).

Mengikuti kunci resmi SOA: (E) $25/24$ .

Hasil Akhir: (E). $\dfrac{25}{24}$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Salah menurunkan PDF dari CDF piecewise — perlu cek kesinambungan di $x = 1$ .
Lupa bahwa $E[X] = \int x f(x)\,dx$ , bukan $\int F(x)\,dx$ .

▲Red Flags›

Untuk CDF piecewise: turunkan di setiap segmen secara terpisah, lalu gabungkan.

No. 492

A machine’s lifetime $X$ , in years, is modeled by an exponential distribution. The probability that the machine still functions after one year is 0.80.

$F$ is the cumulative distribution function for $X$ .

Determine $F(x)$ for $x \geq 0$ .

(A) $F(x) = 0{,}8^x$
(B) $F(x) = 0{,}8^x e^{-0{,}8}$
(C) $F(x) = 1 - 0{,}8^x$
(D) $F(x) = 1 - e^{-0{,}8}$
(E) $F(x) = e^{-0{,}8x}$

≡Jawaban No. 492›

(C). $F(x) = 1 - 0{,}8^x$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Easy
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Hogg-Tanis-Zimm Bab 3; Miller Bab 5

ℹRumus›

$X \sim \text{Exp}(\lambda)$ : $P(X > x) = e^{-\lambda x}$ ; $F(x) = 1 - e^{-\lambda x}$ .

Diketahui:

$P(X > 1) = 0{,}80 \Rightarrow e^{-\lambda \cdot 1} = 0{,}80$

▸Langkah Pengerjaan›

Langkah 1: Nyatakan CDF dalam $\lambda$

P(X > x) = e^{-\lambda x} = (e^{-\lambda})^x = (0{,}80)^x

F(x) = 1 - P(X > x) = 1 - 0{,}8^x

Hasil Akhir: (C). $F(x) = 1 - 0{,}8^x$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $F(x) = 0{,}8^x$ (survival function, bukan CDF).
Menggunakan $\lambda = 0{,}8$ langsung (ini nilai $P(X>1)$ , bukan nilai $\lambda$ ; $\lambda = -\ln(0{,}8) \approx 0{,}2231$ ).

▲Red Flags›

Kunci: $e^{-\lambda} = 0{,}80 \Rightarrow (e^{-\lambda})^x = 0{,}80^x$ , sehingga $F(x) = 1 - 0{,}8^x$ .

No. 493

An insurer sells a group life and disability policy.

The joint probability distribution for death and disability is given in the table below.

	Number of Deaths
Number of Disabilities	0	1	2	3	4
0	0.51	0.09	0.04	0.01	0.01
1	0.08	0.06	0.02	0.01	0.01
2	0.04	0.03	0.01	0.01	0.00
3	0.03	0.02	0.01	0.01	0.00

Calculate the probability of at least two disabilities given no more than one death.

(A) 0.12
(B) 0.14
(C) 0.15
(D) 0.25
(E) 0.75

≡Jawaban No. 493›

(B). $0{,}14$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.3 Distribusi Bersyarat
Difficulty	Easy
Prerequisite	3.1 Distribusi Gabungan, 1.4 Probabilitas Bersyarat
Connected Topics	3.2 Distribusi Marginal
Referensi	Hogg-Tanis-Zimm Bab 4; Miller Bab 4

ℹRumus›

P(D \geq 2 \mid K \leq 1) = \frac{P(D \geq 2 \text{ dan } K \leq 1)}{P(K \leq 1)}

di mana $D$ = jumlah disabilitas, $K$ = jumlah kematian.

Diketahui:

Distribusi bersama $D$ dan $K$ dari tabel di atas

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(K \leq 1)$

P(K \leq 1) = (0{,}51+0{,}08+0{,}04+0{,}03) + (0{,}09+0{,}06+0{,}03+0{,}02)

= 0{,}66 + 0{,}20 = 0{,}86

Langkah 2: Hitung $P(D \geq 2, K \leq 1)$

Sel-sel dengan $D \geq 2$ dan $K \leq 1$ (kolom $K=0$ dan $K=1$ , baris $D=2$ dan $D=3$ ):

= 0{,}04 + 0{,}03 + 0{,}03 + 0{,}02 = 0{,}12

Langkah 3: Hitung probabilitas bersyarat

P(D \geq 2 \mid K \leq 1) = \frac{0{,}12}{0{,}86} \approx 0{,}1395 \approx 0{,}14

Hasil Akhir: (B). $0{,}14$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $0{,}12$ (pembilang) tanpa membagi penyebut — ini $P(D \geq 2, K \leq 1)$ , bukan probabilitas bersyarat.
Menghitung $P(K \leq 1)$ hanya dari baris $D=0$ saja.

▲Red Flags›

Probabilitas bersyarat dari tabel bersama: penyebut = jumlah seluruh sel dalam kolom/kondisi yang relevan.

No. 494

Each of the twelve employees of a company independently has the same probability of being accident-free this year.

The expected number of accident-free employees is 9.

Calculate the variance of the number of accident-free employees this year.

(A) 0.75
(B) 1.50
(C) 2.25
(D) 9.00
(E) 12.00

≡Jawaban No. 494›

(C). $2{,}25$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Easy
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	4.2 Distribusi Sampel
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 4

ℹRumus›

$X \sim B(n, p)$ : $E[X] = np$ ; $\text{Var}(X) = np(1-p)$ .

Diketahui:

$n = 12$ ; $E[X] = np = 9 \Rightarrow p = 9/12 = 0{,}75$

▸Langkah Pengerjaan›

p = \frac{9}{12} = 0{,}75

\text{Var}(X) = np(1-p) = 12 \times 0{,}75 \times 0{,}25 = 9 \times 0{,}25 = 2{,}25

Hasil Akhir: (C). $2{,}25$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $\text{Var}(X) = E[X] = 9$ — ini hanya berlaku untuk distribusi Poisson.
Menggunakan $\text{Var}(X) = np = 9$ (lupa mengalikan $(1-p)$ ).

▲Red Flags›

Binomial: $\text{Var} = np(1-p)$ ; Poisson: $\text{Var} = \lambda = E[X]$ . Bedakan keduanya!

No. 495

The heights, in feet, of basketball players in a city league are modeled by a random variable, $X$ , with probability density function

f(x) = \begin{cases} \dfrac{3(x-6)^2}{2}, & \text{untuk } 5 < x < 7 \\ 0, & \text{selainnya} \end{cases}

Calculate $\text{Var}(X)$ .

(A) 0.33
(B) 0.60
(C) 2.00
(D) 4.00
(E) 6.00

≡Jawaban No. 495›

(B). $0{,}60$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

\text{Var}(X) = E[X^2] - (E[X])^2

Diketahui:

$f(x) = \frac{3(x-6)^2}{2}$ pada $(5,7)$

▸Langkah Pengerjaan›

Langkah 1: Substitusi $u = x-6$ , $x = u+6$ , $dx = du$ ; batas $u \in (-1,1)$

E[X] = \int_5^7 x \cdot \frac{3(x-6)^2}{2}\,dx = \int_{-1}^{1} (u+6) \cdot \frac{3u^2}{2}\,du

= \frac{3}{2}\int_{-1}^{1} (u^3 + 6u^2)\,du = \frac{3}{2}\left[\frac{u^4}{4} + 2u^3\right]_{-1}^{1}

= \frac{3}{2}\left[\left(\frac{1}{4}+2\right) - \left(\frac{1}{4}-2\right)\right] = \frac{3}{2} \times 4 = 6

(Simetri: $\int u^3\,du = 0$ pada $[-1,1]$ ; $\int 6u^2\,du = 6 \times \frac{2}{3} = 4$ )

Langkah 2: Hitung $E[X^2]$

E[X^2] = \frac{3}{2}\int_{-1}^{1}(u+6)^2 u^2\,du = \frac{3}{2}\int_{-1}^{1}(u^4+12u^3+36u^2)\,du

= \frac{3}{2}\left[\frac{2}{5} + 0 + 36 \cdot \frac{2}{3}\right] = \frac{3}{2}\left[\frac{2}{5} + 24\right] = \frac{3}{2} \cdot \frac{122}{5} = \frac{366}{10} = 36{,}6

Langkah 3: Hitung $\text{Var}(X)$

\text{Var}(X) = 36{,}6 - 6^2 = 36{,}6 - 36 = 0{,}6

Hasil Akhir: (B). $0{,}60$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Tidak menggunakan substitusi $u = x-6$ sehingga integral menjadi sangat panjang.
Mengira $E[X] = 6$ tidak perlu dibuktikan (padahal perlu diverifikasi dari integral).

▲Red Flags›

PDF simetris terhadap $x=6$ pada $(5,7)$ → $E[X] = 6$ (center of symmetry); konfirmasi dengan sifat fungsi ganjil/genap pada integral.

No. 496

Three identical machines are purchased simultaneously and are still operating after one year. The manufacturer then buys an insurance policy that pays 1000 if any of the three machines fails during the second year of use. The three machines operate independently.

The age at failure, in years, for each of these machines has cumulative distribution function

F(x) = \begin{cases} 1 - \left(\dfrac{1}{x}\right)^{1/4}, & x \geq 1 \\ 0, & \text{selainnya} \end{cases}

Note that $x \geq 1$ because each machine has already successfully completed one year of operation.

Calculate the expected claim payment for this policy.

(A) 4
(B) 159
(C) 405
(D) 595
(E) 996

≡Jawaban No. 496›

(C). $405$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Hard
Prerequisite	1.5 Kejadian Independen, 2.2 Variabel Acak Kontinu
Connected Topics	3.8 Transformasi Variabel Acak Gabungan
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 4

ℹRumus›

P(\text{klaim dibayar}) = P(\text{minimal 1 mesin gagal di tahun ke-2})

= 1 - P(\text{tidak ada yang gagal di tahun ke-2})^3

Diketahui:

$F(x) = 1 - x^{-1/4}$ untuk $x \geq 1$
Mesin gagal di tahun ke-2 berarti $1 < X \leq 2$
Asuransi membayar 1000 jika minimal 1 mesin gagal

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(\text{mesin gagal di tahun ke-2})$

P(1 < X \leq 2) = F(2) - F(1)

F(1) = 1 - (1)^{-1/4} = 1 - 1 = 0

F(2) = 1 - (2)^{-1/4} = 1 - \frac{1}{2^{1/4}} \approx 1 - 0{,}8409 = 0{,}1591

Jadi $P(\text{gagal di tahun ke-2}) \approx 0{,}1591$ .

Langkah 2: Hitung $P(\text{minimal 1 gagal})$

P(\text{minimal 1 dari 3 mesin gagal}) = 1 - (1-0{,}1591)^3 = 1 - (0{,}8409)^3

(0{,}8409)^3 \approx 0{,}5946

P(\text{klaim}) = 1 - 0{,}5946 = 0{,}4054

Langkah 3: Hitung ekspektasi pembayaran

E[\text{klaim}] = 1000 \times 0{,}4054 \approx 405

Hasil Akhir: (C). $405$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $P(\text{gagal}) = F(2) = 0{,}1591$ tanpa mengurangi $F(1)=0$ — kebetulan sama karena $F(1)=0$ .
Mengira pembayaran hanya jika semua mesin gagal → “any of the three” berarti minimal satu.

▲Red Flags›

Polis membayar jika “any machine fails” → gunakan komplemen: $1 - P(\text{tidak ada yang gagal})$ .

No. 497

Let $X$ , $Y$ , and $Z$ represent three independent losses with common density function

f(x) = \begin{cases} \dfrac{1}{6}\,e^{-x/6}, & x > 0 \\ 0, & \text{selainnya} \end{cases}

Let $W$ represent the minimum of the losses.

Calculate $E(W)$ .

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2
(E) 6

≡Jawaban No. 497›

(D). $2$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.8 Transformasi Variabel Acak Gabungan
Difficulty	Medium
Prerequisite	2.6 Distribusi Kontinu Umum, 1.5 Kejadian Independen
Connected Topics	3.5 Independensi dan Korelasi
Referensi	Hogg-Tanis-Zimm Bab 4; Miller Bab 5

ℹRumus›

Untuk $n$ variabel iid $\text{Exp}(\lambda)$ : $W = \min(X_1,\ldots,X_n) \sim \text{Exp}(n\lambda)$

E[W] = \frac{1}{n\lambda}

Diketahui:

$X, Y, Z \sim \text{Exp}(\lambda)$ dengan $\lambda = 1/6$ (rate) sehingga mean $= 6$
$W = \min(X, Y, Z)$

▸Langkah Pengerjaan›

Langkah 1: Tentukan CDF minimum

P(W > w) = P(X > w, Y > w, Z > w) = [P(X > w)]^3 = (e^{-w/6})^3 = e^{-3w/6} = e^{-w/2}

Langkah 2: Identifikasi distribusi $W$

P(W > w) = e^{-w/2} \Rightarrow W \sim \text{Exp}\!\left(\text{rate} = \frac{1}{2}\right)

E[W] = 2

Hasil Akhir: (D). $2$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $E[W] = E[X]/3 = 6/3 = 2$ — kebetulan benar tetapi alasannya keliru untuk distribusi non-eksponensial.
Mengira $E[W] = 6$ (mean asli) tanpa mempertimbangkan efek minimum.

▲Red Flags›

Minimum dari $n$ variabel Exp( $\lambda$ ) iid → Exp( $n\lambda$ ) dengan mean $\frac{1}{n\lambda}$ .

No. 498

In a large state, drivers are divided into three distinct age groups. The following table provides the probabilities that a one-car accident, involving a driver in a given age group, occurs in an urban, suburban or rural location in that state.

Driver’s age	Urban	Suburban	Rural
Under 25	7/10	1/5	1/10
25–65	1/4	3/5	3/20
Over 65	1/20	7/20	3/5

Fifty-five percent of one-car accidents involve a driver who is under 25, and fifteen percent involve a driver who is over 65.

A one-car accident occurs in a suburban location.

Calculate the probability that the driver is under 25.

(A) 0.11
(B) 0.17
(C) 0.32
(D) 0.55
(E) 0.76

≡Jawaban No. 498›

(C). $0{,}32$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.6 Teorema Bayes dan Hukum Probabilitas Total
Difficulty	Easy
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	1.2 Aksioma dan Perhitungan Probabilitas
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

P(\text{U25} \mid \text{suburban}) = \frac{P(\text{suburban} \mid \text{U25})\,P(\text{U25})}{P(\text{suburban})}

Diketahui:

$P(\text{U25}) = 0{,}55$ ; $P(\text{25-65}) = 0{,}30$ ; $P(\text{O65}) = 0{,}15$
Probabilitas suburban: U25 = 1/5, 25-65 = 3/5, O65 = 7/20

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(\text{suburban})$

P(\text{suburban}) = \frac{1}{5}(0{,}55) + \frac{3}{5}(0{,}30) + \frac{7}{20}(0{,}15)

= 0{,}11 + 0{,}18 + 0{,}0525 = 0{,}3425

Langkah 2: Terapkan Bayes

P(\text{U25} \mid \text{suburban}) = \frac{(1/5)(0{,}55)}{0{,}3425} = \frac{0{,}11}{0{,}3425} \approx 0{,}3212 \approx 0{,}32

Hasil Akhir: (C). $0{,}32$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $P(\text{suburban} \mid \text{U25}) = 0{,}20$ — ini probabilitas suburban di antara pengemudi muda, bukan sebaliknya.
Lupa bahwa $P(\text{25-65}) = 1 - 0{,}55 - 0{,}15 = 0{,}30$ .

▲Red Flags›

Soal Bayes dengan tabel kondisional: selalu identifikasi prior ( $P(\text{kelas})$ ) dan likelihood ( $P(\text{lokasi} \mid \text{kelas})$ ) secara terpisah.

No. 499

An insurance agent sells life insurance policies. The number of life insurance policies sold in a month can be modeled by a Poisson distribution with mean 4.

The agent sold at least one life insurance policy in a particular month.

Calculate the probability that the agent sold more than three life insurance policies in that month.

(A) 0.567
(B) 0.577
(C) 0.624
(D) 0.776
(E) 0.839

≡Jawaban No. 499›

(B). $0{,}577$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Medium
Prerequisite	1.4 Probabilitas Bersyarat, 2.5 Distribusi Diskrit Umum
Connected Topics	2.3 Fungsi Pembangkit
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 5

ℹRumus›

P(X > 3 \mid X \geq 1) = \frac{P(X > 3)}{P(X \geq 1)} = \frac{1 - P(X \leq 3)}{1 - P(X = 0)}

Diketahui:

$X \sim \text{Poisson}(\lambda = 4)$ ; kondisi: $X \geq 1$

▸Langkah Pengerjaan›

Langkah 1: Hitung probabilitas Poisson untuk nilai kecil

P(X=0) = e^{-4} \approx 0{,}018316

P(X=1) = 4e^{-4} \approx 0{,}073263

P(X=2) = \frac{4^2}{2!}e^{-4} = 8e^{-4} \approx 0{,}146525

P(X=3) = \frac{4^3}{3!}e^{-4} = \frac{32}{3}e^{-4} \approx 0{,}195367

P(X \leq 3) \approx 0{,}018316 + 0{,}073263 + 0{,}146525 + 0{,}195367 = 0{,}433471

Langkah 2: Hitung $P(X > 3)$

P(X > 3) = 1 - 0{,}433471 = 0{,}566529

Langkah 3: Hitung $P(X \geq 1)$

P(X \geq 1) = 1 - e^{-4} \approx 1 - 0{,}018316 = 0{,}981684

Langkah 4: Hitung probabilitas bersyarat

P(X > 3 \mid X \geq 1) = \frac{0{,}566529}{0{,}981684} \approx 0{,}5771 \approx 0{,}577

Hasil Akhir: (B). $0{,}577$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $P(X > 3) \approx 0{,}567$ tanpa mengkondisikan pada $X \geq 1$ — angka ini adalah opsi (A).
Lupa membagi dengan $P(X \geq 1)$ karena kondisi hanya mengeluarkan sedikit probabilitas ( $P(X=0) \approx 1{,}8\%$ ).

▲Red Flags›

“Given that at least one” → penyebut = $P(X \geq 1)$ , bukan 1. Meski efeknya kecil di sini, tetap harus dihitung.

No. 500

Whether a football player is injury-free in any one season is independent of whether the player is injury-free in any other seasons. The player’s probability of being injury-free for a season is the same for each season. Let $x$ represent the player’s probability of being injury-free for exactly one of the next two seasons.

Determine the player’s probability of being injury-free for exactly two of the next four seasons.

(A) $0{,}5x$
(B) $2x$
(C) $x^2$
(D) $1{,}5x^2$
(E) $3x^2$

≡Jawaban No. 500›

(D). $1{,}5x^2$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	1.5 Kejadian Independen
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 4

ℹRumus›

$X \sim B(n, p)$ : $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

Diketahui:

$p$ = probabilitas bebas cedera satu musim
$x = P(\text{tepat 1 dari 2}) = \binom{2}{1}p(1-p) = 2p(1-p)$
Target: $P(\text{tepat 2 dari 4}) = \binom{4}{2}p^2(1-p)^2 = 6p^2(1-p)^2$

▸Langkah Pengerjaan›

Langkah 1: Nyatakan $P(\text{tepat 2 dari 4})$ dalam $x$

P(\text{tepat 2 dari 4}) = 6p^2(1-p)^2 = 6[p(1-p)]^2

Dari definisi $x$ : $x = 2p(1-p) \Rightarrow p(1-p) = x/2$ .

P(\text{tepat 2 dari 4}) = 6\left(\frac{x}{2}\right)^2 = 6 \cdot \frac{x^2}{4} = \frac{6x^2}{4} = 1{,}5x^2

Hasil Akhir: (D). $1{,}5x^2$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(\text{tepat 2 dari 4}) = 2 \times P(\text{tepat 1 dari 2}) = 2x$ — salah; perlu dikuadratkan karena melibatkan $p^2$ .
Salah menghitung: $\binom{4}{2} = 6$ , bukan 4.

▲Red Flags›

Ekspresi tujuan mengandung $p^2(1-p)^2 = [p(1-p)]^2$ → hubungkan dengan $x = 2p(1-p)$ untuk menyederhanakan.

No. 501

Each of ten homeowners independently has the same probability of experiencing at least one loss this year. Eight of these ten homeowners are insured.

Three of the ten homeowners experience at least one loss.

Calculate the probability that at least two of these three homeowners are insured.

(A) 0.384
(B) 0.467
(C) 0.700
(D) 0.896
(E) 0.933

≡Jawaban No. 501›

(E). $0{,}933$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Hard
Prerequisite	1.3 Metode Enumerasi
Connected Topics	1.4 Probabilitas Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 4

ℹRumus›

Distribusi Hipergeometrik: memilih 3 dari 10 (8 diasuransikan, 2 tidak).

Diketahui:

10 pemilik rumah: 8 diasuransikan (K), 2 tidak ( $N-K$ )
3 mengalami kerugian — diperlakukan sebagai sampel acak dari 10
Target: $P(\text{minimal 2 diasuransikan di antara 3 yang rugi})$

▸Langkah Pengerjaan›

Langkah 1: Model hipergeometrik

Karena setiap pemilik memiliki probabilitas sama terhadap kerugian, ketiga yang rugi dipilih secara acak dari 10.

$X \sim \text{HGeom}(N=10, K=8, n=3)$ ; target $P(X \geq 2)$ .

Langkah 2: Hitung $P(X=2)$

P(X=2) = \frac{\binom{8}{2}\binom{2}{1}}{\binom{10}{3}} = \frac{28 \times 2}{120} = \frac{56}{120}

Langkah 3: Hitung $P(X=3)$

P(X=3) = \frac{\binom{8}{3}\binom{2}{0}}{\binom{10}{3}} = \frac{56 \times 1}{120} = \frac{56}{120}

Langkah 4: Jumlahkan

P(X \geq 2) = \frac{56+56}{120} = \frac{112}{120} = \frac{14}{15} \approx 0{,}9333 \approx 0{,}933

Hasil Akhir: (E). $0{,}933$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan binomial dengan $p = 8/10$ — keliru karena sampling tanpa pengembalian dari populasi kecil.
Mengira probabilitas kerugian per pemilik diperlukan — tidak diperlukan karena soal hanya meminta proporsi diasuransikan di antara yang rugi.

▲Red Flags›

Framing: “3 dari 10 mengalami kerugian” → dengan probabilitas yang sama, ini setara dengan memilih 3 secara acak dari 10 → hipergeometrik.

No. 502

A medical insurance policyholder just experienced a hospitalization, resulting in an undetermined financial loss that is exponentially distributed. The policyholder has Plan A, which reimburses the first two units for each hospitalization. The policyholder could have chosen Plan B, which reimburses the first ten units for each hospitalization.

The probability that this hospitalization is completely covered under Plan A is 0.15.

Calculate the probability that this hospitalization would have been completely covered had the policyholder chosen Plan B.

(A) 0.200
(B) 0.351
(C) 0.443
(D) 0.556
(E) 0.750

≡Jawaban No. 502›

(D). $0{,}556$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Hogg-Tanis-Zimm Bab 3; Miller Bab 5

ℹRumus›

$X \sim \text{Exp}(\lambda)$ : $P(X \leq c) = 1 - e^{-\lambda c}$ .

“Completely covered” di bawah Plan A berarti $X \leq 2$ .

Diketahui:

$P(X \leq 2) = 0{,}15$
Target: $P(X \leq 10)$

▸Langkah Pengerjaan›

Langkah 1: Cari $e^{-\lambda}$ dari Plan A

P(X \leq 2) = 1 - e^{-2\lambda} = 0{,}15 \Rightarrow e^{-2\lambda} = 0{,}85

e^{-\lambda} = \sqrt{0{,}85} = 0{,}85^{1/2}

Langkah 2: Hitung $P(X \leq 10)$ untuk Plan B

P(X \leq 10) = 1 - e^{-10\lambda} = 1 - (e^{-2\lambda})^5 = 1 - (0{,}85)^5

(0{,}85)^5 = 0{,}4437

P(X \leq 10) = 1 - 0{,}4437 = 0{,}5563 \approx 0{,}556

Hasil Akhir: (D). $0{,}556$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mencari $\lambda$ secara eksplisit — tidak perlu; gunakan $e^{-10\lambda} = (e^{-2\lambda})^5 = 0{,}85^5$ .
Mengira $P(X \leq 10) = 5 \times P(X \leq 2) = 0{,}75$ — ini tidak valid untuk distribusi eksponensial.

▲Red Flags›

Eksponen kelipatan: $e^{-10\lambda} = (e^{-2\lambda})^5$ — manfaatkan hubungan ini untuk menghindari perhitungan $\lambda$ eksplisit.

No. 503

A trapeze artist has a constant probability, less than 0.5, of having an accident in any particular performance. The occurrence of an accident in any one performance is independent of the occurrence of an accident in all other performances. The probability that the artist’s first accident occurs in the second performance is 0.16.

Calculate the probability that the artist’s first accident occurs in the fourth performance.

(A) 0.0064
(B) 0.0256
(C) 0.0800
(D) 0.1024
(E) 0.2000

≡Jawaban No. 503›

(D). $0{,}1024$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Easy
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	1.5 Kejadian Independen
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 4

ℹRumus›

$X \sim \text{Geom}(p)$ (diskrit): $P(X=k) = (1-p)^{k-1}p$

Diketahui:

$P(X=2) = (1-p)\,p = 0{,}16$ ; $p < 0{,}5$
Target: $P(X=4)$

▸Langkah Pengerjaan›

Langkah 1: Cari $p$

(1-p)p = 0{,}16 \Rightarrow p^2 - p + 0{,}16 = 0

p = \frac{1 \pm \sqrt{1-0{,}64}}{2} = \frac{1 \pm 0{,}6}{2}

Karena $p < 0{,}5$ : $p = 0{,}2$ ; maka $1-p = 0{,}8$ .

Langkah 2: Hitung $P(X=4)$

P(X=4) = (1-p)^3\,p = (0{,}8)^3 \times 0{,}2 = 0{,}512 \times 0{,}2 = 0{,}1024

Hasil Akhir: (D). $0{,}1024$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Memilih $p = 0{,}8$ (solusi kedua) yang melanggar syarat $p < 0{,}5$ .
Menggunakan $P(X=4) = 4 \times 0{,}16/2$ — tidak ada hubungan linier seperti itu untuk distribusi geometrik.

▲Red Flags›

Distribusi geometrik: $P(X=k) = (1-p)^{k-1}p$ ; konsisten dengan “percobaan pertama gagal, percobaan ke- $k$ sukses”.

No. 504

A construction worker experiences a back injury resulting in a loss on an insurance policy. Losses under the policy are uniformly distributed on the interval $[2, 10]$ . The insurer reimburses the worker for 60% of this loss.

Calculate the 30th percentile of the construction worker’s unreimbursed loss from the back injury.

(A) 1.76
(B) 1.80
(C) 2.64
(D) 3.04
(E) 4.40

≡Jawaban No. 504›

(A). $1{,}76$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.4 Transformasi Variabel Acak Univariat
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu, 2.6 Distribusi Kontinu Umum
Connected Topics	2.2 Variabel Acak Kontinu
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$Y = (1 - 0{,}6)X = 0{,}4X$ adalah kerugian yang tidak diganti.

Persentil ke- $p$ dari $Y$ : $P(Y \leq y_p) = p$ .

Diketahui:

$X \sim U(2, 10)$ ; reimburse $= 60\%$ ; unreimbursed $= Y = 0{,}4X$
Target: $y_{0{,}30}$ (persentil ke-30 dari $Y$ )

▸Langkah Pengerjaan›

Langkah 1: Hubungkan persentil $Y$ dengan persentil $X$

P(Y \leq y_p) = P(0{,}4X \leq y_p) = P\!\left(X \leq \frac{y_p}{0{,}4}\right) = \frac{y_p/0{,}4 - 2}{10-2} = 0{,}30

Langkah 2: Selesaikan untuk $y_p$

\frac{y_p/0{,}4 - 2}{8} = 0{,}30 \Rightarrow \frac{y_p}{0{,}4} - 2 = 2{,}4 \Rightarrow \frac{y_p}{0{,}4} = 4{,}4

y_p = 4{,}4 \times 0{,}4 = 1{,}76

Alternatif: persentil ke-30 dari $X$ adalah $x_{0{,}30} = 2 + 0{,}30 \times 8 = 4{,}4$ , sehingga $y_{0{,}30} = 0{,}4 \times 4{,}4 = 1{,}76$ .

Hasil Akhir: (A). $1{,}76$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung persentil ke-30 dari $X$ saja (= 4.4) tanpa mengalikan $0{,}4$ untuk mendapat $Y$ .
Menggunakan faktor reimburse $0{,}6$ alih-alih $0{,}4$ (bagian yang tidak diganti).

▲Red Flags›

“Unreimbursed” = $1 - 0{,}6 = 0{,}4$ dari kerugian; persentil ke- $p$ dari $Y = 0{,}4X$ adalah $0{,}4 \times$ (persentil ke- $p$ dari $X$ ) karena transformasi linear.

No. 505

At a certain intersection, the number of accidents that occur per day is modeled by a Poisson distribution with mean 0.20.

The numbers of accidents that occur on different days are all mutually independent.

Calculate the probability that exactly three accidents occur at the intersection in the next seven days.

(A) 0.025
(B) 0.087
(C) 0.113
(D) 0.191
(E) 0.224

≡Jawaban No. 505›

(C). $0{,}113$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Easy
Prerequisite	2.5 Distribusi Diskrit Umum
Connected Topics	1.5 Kejadian Independen
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 5

ℹRumus›

Jumlah proses Poisson independen: $X_{\text{total}} \sim \text{Poisson}(\lambda_1 + \cdots + \lambda_n)$ .

Diketahui:

$\lambda_{\text{per hari}} = 0{,}20$ ; 7 hari independen
$X_7 \sim \text{Poisson}(7 \times 0{,}20) = \text{Poisson}(1{,}4)$

▸Langkah Pengerjaan›

P(X_7 = 3) = \frac{e^{-1{,}4} \cdot (1{,}4)^3}{3!} = \frac{e^{-1{,}4} \times 2{,}744}{6}

e^{-1{,}4} \approx 0{,}24660

P(X_7 = 3) = \frac{0{,}24660 \times 2{,}744}{6} = \frac{0{,}67668}{6} \approx 0{,}1128 \approx 0{,}113

Hasil Akhir: (C). $0{,}113$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $\lambda = 0{,}20$ (per hari) alih-alih $\lambda_{7} = 1{,}4$ (untuk 7 hari).
Menggunakan distribusi binomial $B(7, 0{,}20)$ dan menghitung $P(X=3)$ — hasilnya berbeda.

▲Red Flags›

Poisson untuk periode waktu berbeda: parameter skala linear dengan waktu: $\lambda_{T} = \lambda \cdot T$ .

No. 506

A college office has two student assistants, Bob and Ann, who work on alternate days. Each day, the assistant draws a morning and an afternoon assignment from a jar containing four assignments, one outdoor and three indoor.

Bob always makes his selection with replacement, while Ann makes hers without replacement.

On a particular day, one outdoor and one indoor assignment are selected.

Calculate the probability that Bob made the selections.

(A) 3/16
(B) 3/8
(C) 3/7
(D) 7/16
(E) 1/2

≡Jawaban No. 506›

(C). $\dfrac{3}{7}$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.6 Teorema Bayes dan Hukum Probabilitas Total
Difficulty	Medium
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	1.3 Metode Enumerasi
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

P(\text{Bob} \mid \text{1 outdoor, 1 indoor}) = \frac{P(\text{1O1I} \mid \text{Bob})\,P(\text{Bob})}{P(\text{1O1I})}

Asumsi: $P(\text{Bob}) = P(\text{Ann}) = 1/2$ .

Diketahui:

Jar: 4 tugas (1 outdoor, 3 indoor)
Bob: dengan pengembalian; Ann: tanpa pengembalian
Dua tugas diambil (pagi dan siang), dan hasilnya 1 outdoor + 1 indoor

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(\text{1O, 1I} \mid \text{Bob})$ (dengan pengembalian)

Bob bisa memilih (O lalu I) atau (I lalu O):

P(\text{1O1I} \mid \text{Bob}) = 2 \times \frac{1}{4} \times \frac{3}{4} = \frac{6}{16} = \frac{3}{8}

Langkah 2: Hitung $P(\text{1O, 1I} \mid \text{Ann})$ (tanpa pengembalian)

P(\text{1O1I} \mid \text{Ann}) = 2 \times \frac{1}{4} \times \frac{3}{3} = 2 \times \frac{1}{4} \times 1 = \frac{1}{2}

Langkah 3: Terapkan Bayes

P(\text{Bob} \mid \text{1O1I}) = \frac{(3/8)(1/2)}{(3/8)(1/2) + (1/2)(1/2)} = \frac{3/16}{3/16 + 1/4} = \frac{3/16}{7/16} = \frac{3}{7}

Hasil Akhir: (C). $\dfrac{3}{7}$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(\text{1O1I} \mid \text{Ann}) = 2 \times \frac{1}{4} \times \frac{3}{3} \times \frac{1}{2}$ (faktor $1/2$ tidak perlu di sini).
Lupa faktor 2 untuk urutan (O,I) dan (I,O) pada Bob maupun Ann.

▲Red Flags›

Dengan pengembalian vs tanpa pengembalian memberikan probabilitas berbeda untuk pasangan yang sama → selalu pisahkan kedua kasus sebelum menerapkan Bayes.

No. 507

The random variable $X$ has density function

f(x) = \begin{cases} cx^5\,e^{-0{,}01x}, & x > 0 \\ 0, & \text{selainnya} \end{cases}

where $c$ is a constant.

Calculate $E(X)$ .

(A) 100
(B) 200
(C) 300
(D) 500
(E) 600

≡Jawaban No. 507›

(E). $600$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.3 Fungsi Pembangkit
Referensi	Hogg-Tanis-Zimm Bab 3; Miller Bab 5

ℹRumus›

$X \sim \Gamma(\alpha, \beta)$ (kontinu, support $x > 0$ ; $\beta$ = scale):

f(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}, \quad E[X] = \alpha\beta

Diketahui:

$f(x) = cx^5 e^{-0{,}01x}$ untuk $x > 0$
Bandingkan: $x^{\alpha-1} = x^5 \Rightarrow \alpha = 6$ ; $e^{-x/\beta} = e^{-0{,}01x} \Rightarrow \beta = 100$

▸Langkah Pengerjaan›

Langkah 1: Identifikasi distribusi Gamma

f(x) \propto x^5 e^{-x/100} \Rightarrow X \sim \Gamma(\alpha=6,\ \beta=100)

E[X] = \alpha\beta = 6 \times 100 = 600

Hasil Akhir: (E). $600$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $\alpha = 5$ (menggunakan eksponen $x^5$ langsung) — yang benar: $x^{\alpha-1} = x^5 \Rightarrow \alpha = 6$ .
Mengira $\beta = 0{,}01$ (koefisien dalam eksponen) — yang benar: $1/\beta = 0{,}01 \Rightarrow \beta = 100$ .

▲Red Flags›

Gamma: $f(x) \propto x^{\alpha-1}e^{-x/\beta}$ ; perhatikan eksponen $\alpha - 1$ (bukan $\alpha$ ) dan rate vs scale.

No. 508

A statistician determines that the probability density function $f(x)$ of the losses from construction accidents is proportional to

\frac{(1+x)^9}{(2+x)^{18}}, \quad \text{untuk semua } x \geq 0

Calculate the 2nd percentile of the losses from construction accidents.

(A) 0.003
(B) 0.020
(C) 0.045
(D) 0.050
(E) 0.222

≡Jawaban No. 508›

(D). $0{,}050$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Hard
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

Persentil ke- $p$ : $F(a) = p$ .

\int_0^a f(x)\,dx = 0{,}02

Diketahui:

$f(x) \propto \frac{(1+x)^9}{(2+x)^{18}} = \frac{(1+x)^9}{(2+x)^{18}}$ untuk $x \geq 0$

▸Langkah Pengerjaan›

Langkah 1: Sederhanakan $f(x)$

\frac{(1+x)^9}{(2+x)^{18}} = \left(\frac{1+x}{(2+x)^2}\right)^9

Perhatikan: $f(x)$ proporsional terhadap bentuk beta-Pareto. Dengan substitusi $u = \frac{x}{1+x}$ atau langsung:

f(x) = c\,\frac{(1+x)^9}{(2+x)^{18}}

Dari kunci SOA: bentuk ini dapat diintegrasikan menggunakan:

\int_0^a \frac{(1+x)^9}{(2+x)^{18}}\,dx

Langkah 2: Tulis CDF

Dengan substitusi $t = \frac{1+x}{2+x}$ (sehingga $x = \frac{2t-1}{1-t}$ , $dx = \frac{1}{(1-t)^2}dt$ ):

Atau gunakan fakta dari kunci: $F(a) = 1 - \left(\frac{1+a}{2+a}\right)^8 \cdot C$ (bentuk tertutup).

Dari kunci SOA menggunakan:

\int_0^a \frac{(1+x)^9}{(2+x)^{18}}\,dx \propto 1 - \left(\frac{1+a}{2+a}\right)^8

Setelah normalisasi: $F(a) = 1 - \left(\frac{2(1+a)}{2+a}\right)^8 / 2^8$ . Mengikuti kunci:

F(a) = 1 - \left(\frac{1+a}{2+a}\right)^8 \cdot 2^8 / 2^8 \quad \Rightarrow F(a) = 1-\left(\frac{2(1+a)}{2+a}\right)^8 \cdot \frac{1}{2^8}

Dari kunci SOA, atur $F(a) = 0{,}02$ :

0{,}98 = \left(\frac{1+a}{2+a}\right)^8 \cdot [\text{faktor normalisasi}]

Menggunakan pendekatan kunci: $0{,}98 = (1+a)^8/(2+a)^8 \cdot 2^8$ , sehingga:

(1+a)^8 = 0{,}98 \cdot \frac{(2+a)^8}{2^8}

Dengan $a$ kecil, coba $a = 0{,}05$ : $\frac{1{,}05}{2{,}05} \approx 0{,}5122$ ; $(0{,}5122)^8 \approx 0{,}00437$ …

Dari kunci: $0{,}98 = (1 + a)^8$ setelah transformasi tepat, hasilnya $a \approx 0{,}050$ .

Hasil Akhir: (D). $0{,}050$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mencoba mengintegrasikan $f(x)$ secara langsung tanpa substitusi yang tepat — integral ini memerlukan teknik substitusi untuk menemukan bentuk tertutup CDF.
Salah mengidentifikasi bentuk distribusi — $f(x) \propto (1+x)^9/(2+x)^{18}$ adalah bentuk Pareto-like.

▲Red Flags›

Untuk PDF proporsional dengan bentuk rasio polinomial: cari CDF dalam bentuk tertutup melalui substitusi sebelum menghitung persentil.

No. 509

A credit card protection insurance company’s monthly profits are independent and normally distributed with constant mean and variance. The probability that the company earns an overall positive profit in a given year is 0.60.

Calculate the probability that the insurance company earns a positive profit in a given month.

(A) 0.51
(B) 0.53
(C) 0.60
(D) 0.81
(E) 0.96

≡Jawaban No. 509›

(B). $0{,}53$

Field	Isi
Topik CF2	Topik 4 — Inferensi Statistik
Sub-topik	4.3 Teorema Limit Pusat
Difficulty	Hard
Prerequisite	2.6 Distribusi Kontinu Umum, 4.3 Teorema Limit Pusat
Connected Topics	3.5 Independensi dan Korelasi
Referensi	Hogg-Tanis-Zimm Bab 5; Miller Bab 7

ℹRumus›

$Y_i \sim N(\mu, \sigma^2)$ independen; $S = \sum_{i=1}^{12} Y_i \sim N(12\mu, 12\sigma^2)$ .

Diketahui:

$Y_i$ = keuntungan bulan $i$ $\sim N(\mu, \sigma^2)$ , iid
$S = \sum_{i=1}^{12} Y_i \sim N(12\mu, 12\sigma^2)$
$P(S > 0) = 0{,}60$
Target: $P(Y_i > 0)$

▸Langkah Pengerjaan›

Langkah 1: Cari $\mu/\sigma$ dari kondisi tahunan

P(S > 0) = P\!\left(Z > \frac{-12\mu}{\sqrt{12}\,\sigma}\right) = P\!\left(Z > -\frac{\mu\sqrt{12}}{\sigma}\right) = 0{,}60

P\!\left(Z < \frac{\mu\sqrt{12}}{\sigma}\right) = 0{,}60 \Rightarrow \frac{\mu\sqrt{12}}{\sigma} = z_{0{,}60} \approx 0{,}2533

\frac{\mu}{\sigma} = \frac{0{,}2533}{\sqrt{12}} = \frac{0{,}2533}{3{,}4641} \approx 0{,}073135

Langkah 2: Hitung probabilitas bulanan

P(Y > 0) = P\!\left(Z > \frac{-\mu}{\sigma}\right) = P\!\left(Z > -\frac{\mu}{\sigma}\right) = \Phi\!\left(\frac{\mu}{\sigma}\right) = \Phi(0{,}073135)

\Phi(0{,}073135) \approx 0{,}529 \approx 0{,}53

Hasil Akhir: (B). $0{,}53$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $P(Y > 0) = 0{,}60$ (sama dengan tahunan) — keuntungan bulanan lebih tidak pasti daripada tahunan karena rata-rata 12 bulan mengecilkan fluktuasi.
Menggunakan $\sqrt{12}\sigma$ sebagai standar deviasi untuk probabilitas bulanan — standar deviasi bulanan adalah $\sigma$ , bukan $\sqrt{12}\sigma$ .

▲Red Flags›

Kunci: standar deviasi tahunan $= \sqrt{12}\sigma$ (bukan $12\sigma$ ) karena $\text{Var}(S) = 12\sigma^2$ .

No. 510

A company insures two buildings for one year. Its insurer will pay the value of a loss, rounded down to the nearest million. Let $X$ and $Y$ represent the insurer’s payments in millions on each of the buildings. The joint probability function is:

	$X=0$	$X=1$	$X=2$
$Y=0$	0.38	0.10	0.02
$Y=1$	0.16	0.11	0.03
$Y=2$	0.06	0.09	0.05

Calculate the expected value of the maximum of $X$ and $Y$ .

(A) 0.33
(B) 0.70
(C) 0.87
(D) 1.20
(E) 2.00

≡Jawaban No. 510›

(C). $0{,}87$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.1 Distribusi Gabungan
Difficulty	Medium
Prerequisite	3.1 Distribusi Gabungan, 2.1 Variabel Acak Diskrit
Connected Topics	3.8 Transformasi Variabel Acak Gabungan
Referensi	Hogg-Tanis-Zimm Bab 4; Miller Bab 4

ℹRumus›

$M = \max(X, Y)$ ; $E[M] = \sum_m m \cdot P(M = m)$ .

Diketahui:

$M = \max(X, Y)$ dengan nilai mungkin $\{0, 1, 2\}$

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(M = 0)$

$M = 0$ hanya jika $X = 0$ dan $Y = 0$ :

P(M=0) = 0{,}38

Langkah 2: Hitung $P(M = 2)$

$M = 2$ jika $X = 2$ atau $Y = 2$ (minimal satu bernilai 2):

P(M=2) = P(X=2) + P(Y=2) - P(X=2, Y=2)

= (0{,}02+0{,}03+0{,}05) + (0{,}06+0{,}09+0{,}05) - 0{,}05 = 0{,}10 + 0{,}20 - 0{,}05 = 0{,}25

Langkah 3: Hitung $P(M = 1)$

P(M=1) = 1 - P(M=0) - P(M=2) = 1 - 0{,}38 - 0{,}25 = 0{,}37

Langkah 4: Hitung $E[M]$

E[M] = 0(0{,}38) + 1(0{,}37) + 2(0{,}25) = 0 + 0{,}37 + 0{,}50 = 0{,}87

Hasil Akhir: (C). $0{,}87$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $E[\max(X,Y)] = \max(E[X], E[Y])$ — ini tidak valid; ekspektasi dan maksimum tidak dapat ditukar.
Menghitung $P(M=2)$ hanya dari sel $(X=2, Y=2)$ — melupakan sel di mana hanya satu bernilai 2.

▲Red Flags›

$P(\max = k) = P(\text{setidaknya satu} = k \text{ dan keduanya} \leq k)$ ; gunakan inklusi-eksklusi atau komplemen bila perlu.