Question

A die is thrown three times,

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.
Determine 36P where P(E|F)

Answer 1

A die is thrown three times P(E|F) is 0.16.

If a die is thrown three times, then the number of elements in the sample space will be 6 * 6 * 6 = 216

E=\ (1, 1, 4), (1, 2, 4),...(1,6,4) (2, 1, 4), (2, 2, 4), ... (2, 6, 4)

(3, 1, 4), (3, 2, 4),...(3,6,4)

(4, 1, 4), (4, 2, 4),...(4,6,4)

(5, 1, 4), (5, 2, 4), ... (5, 6, 4)

(6, 1, 4), (6, 2, 4),...(6,6,4)\

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

E cap F = \{(6, 5, 4)\}

P(F)= 6/216 * and P(E cap F)= 1 216

.: P(E|F)= P(E cap F) P(F) = (1/216)/(6/216) = 1/6 = 0.16