NCERT Solutions
Class 12 Maths
Probability

Ex.13.1 Q.8
A die is thrown three times,
E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses
If a die is thrown three times, then the number of elements in the sample space will be
6 × 6 × 6 = 216 i.e.
(1, 1, 4), (1, 2, 4), (1, 1, 4), ……………….., (1, 6, 4)
(1, 1, 4), (2, 2, 4), (2, 3, 4), ……………….., (2, 6, 4)
E = (3, 1, 4), (3, 2, 4), (3, 3, 4), ……………….., (3, 6, 4)
(4, 1, 4), (4, 2, 4), (4, 3, 4), ……………….., (4, 5, 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)}
So, E ∩ F = {(6, 5, 4)}
Now, P(F) and P(E ∩ F) =
So