Question

A balanced die is thrown three times. Let A be the event that 4 appears on the third toss and B be the event that 6 and 5 appears respectively on first two tosses. Find P(A | B).

Answer 1

Answer:

P(A | B) = 1/6

Step-by-step explanation:

Dice has 6 numbers 1 to 6

Each number has equal probability to appear in a throw

P(A | B) = P(A∩B)/P(B)

A∩B = 4 appears on the third toss & 6 and 5 appears respectively on first two tosses

P(A∩B) = (1/6)(1/6)(1/6) = 1/216

P(B) = (1/6)(1/6) *1 = 1/36

P(A | B) = (1/216)/(1/36) = 1/6

Answer 2

Answer:

P(A | B) = 1/6

Step-by-step explanation:

Dice has 6 numbers 1 to 6

Each number has equal probability to appear in a throw

P(A | B) = P(A∩B)/P(B)

A∩B = 4 appears on the third toss & 6 and 5 appears respectively on first two tosses

P(A∩B) = (1/6)(1/6)(1/6) = 1/216

P(B) = (1/6)(1/6) *1 = 1/36

P(A | B) = (1/216)/(1/36) = 1/6