Jawaban Soal 1 (Peluang Kejadian Majemuk) Bagian 1

1.  a. A = {2,4,6} → n(A) = 3

    B = {2,3,5} → n(B) = 3

    S = {1,2,3,4,5,6} → n(S) = 6

    A ∩ B = {2} → n(A ∩ B) = 1

 P(A ᴜ B)      = P(A) + P(B) – P(A ∩ B)

              = n(A)/n(S)  +  n(B)/n(S) – n(A ∩ B)/n(S)

              = 3/6 + 3/6 – 1/6

              = 5/6

b. Misalkan C : kejadian munculnya bilangaan prima ganjil

C = {3,5} → n(C) = 2

A ∩ C = {} → Kejaian saling lepas

P (A ᴜ C)                                                                                                              = P(A) + P(C)

              =  3/6 + 2/6

              = 5/6

2. S = {1,2,3,4,5,6,7,8} → n(S) = 8

a. A = kejadian terambilnya nomor 8 = {8} → n(A) = 1

P(A) = n(A)/n(S) = 1/8

b. P(A^c) = 1 – P(A) = 1 – 1/8 = 7/8

c. Misalkan,

A : kejadian terambilnya bola nomor genap = {2,4,6,8} → n(A) = 4

B : kejadian terambilnya bola nomor  = {7} → n(B) = 1

(P(A ᴜ B) = P(A) + P(B) → Kejadian saling lepas

              = 4/8 + 1/8 = 5/8

3. Misalkan,

A : jumlah siswa gemar voli

B : jumlah siswa gemar tenis

Maka,

n(A) = 22

n(B) = 17

n(A ᴜ B)^c = 4

n(S) = 36

a. P(A) = 22/36 = 11/18

b. P(B) = 17/36

Untuk bagian c, d, e, dan f.

 n(S) – n(A ᴜ B)^c = n(A) + n(B) – n(A∩B)

    36 – 4 = 22 + 17 - n(A∩B)

    n(A∩B) = 39 – 32 = 7

 

c. P(A ᴜ B) = P(A) + P(B) – P(A∩B)

                = 22/35 + 17/36 – 7/36 = 32/36 = 8/9

d. P(hanya gemar olahraga voli) = 15/36 = 5/12

e. P(hanya gemar olahraga tenis) = 10/36 = 5/18

f. P (A∩B) = 7/36

Semoga bermanfaat!!!