Question

Determine P(E/F). A die is thrown three times, E: ‘4 appears on the third toss’, F: ‘6 and 5 appears respectively on the two tosses’.

Answer 1

n(S) = 6³ = 216

E = {( 1, 1, 4), (1, 2, 4), (1, 3, 4)……….(1, 6, 4),

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

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

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

(5, 1, 4), (5, 2, 4), (5, 3, 4)……..(5, 6, 4),

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

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

⇒ E ∩ F = {(6, 5, 4)}

P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)

Then, P(E/F) = \(\frac{P(E ∩ F)}{P(F)} = \frac{\frac{1}{216}}{{\frac{6}{216}}} = \frac{1}{6}\)