Jawab:

1. Pelemparan uang logam

S = {A,G}

n(S) = 2

n = 70 kali

A = Peluang munculnya sisi gambar

A = {G)

n(A) = 1

Frekuensi Harapan?

Fh = P(A) x n

Fh = n(A)/n(S) x n

Fh = 1/2 x 70

Fh = 35 kali

2. Pelemparan dua buah dadu

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

n(S) = 36

n = 120 kali

A = Peluang Munculnya mata dadu berjumlah tiga

A = {(1,2), (2,1)}

n(A) = 2

Frekuensi harapan?

Fh = P(A) x n

Fh = n(A)/n(S) x n

Fh = 2/36 x 120

Fh = 120/18

Fh = 6,67 kali

2. Pelemparan dua buah dadu

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

n(S) = 36

Peluang Munculnya mata dadu jumlah kedua mata dadu 10 atau 5.

Merupakan peluang kejadian saling lepas:

P(A∪B) = P(A) + P(B)

A = Peluang Munculnya mata dadu jumlah kedua mata dadu 10

B = Peluang Munculnya mata dadu jumlah kedua mata dadu 5

A = {(4,6), (6,4), (5,5)}

n(A) = 3

B = {(1,4), (4,1), (2,3), (3,2)}

n(B) = 4

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

P(A) = 3/36

P(B) = n(B)/n(S)

P(B) = 4/36

Peluang Munculnya mata dadu jumlah kedua mata dadu 10 atau 5 yaitu :

P(A∪B) = P(A) + P(B)

P(A∪B) = 3/36 + 4/36

P(A∪B) = 7/36