Question

Two dice are thrown simultaneously. What is the probability that (i) 5 will not come up on either of them? (ii) 5 will come up on at least one of them? (iii) 5 will come up at both the dice?

Solution

When two dice are thrown simultaneously, the possible outcomes are: (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) Number of all possible outcomes  = 36 (i) Let E1 be the event in which 5 will not come up on either of them.     Then, the favourable outcomes are:     (1,1), (1,2), (1,3), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,6), (3,1), (3,2), (3,3), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4),  (4,6), (6,1), (6,2), (6,3), (6,4) and (6,6).     Number of favourable outcomes = 25 ∴ P (5 will not come up on either of them) = P (E1) = $\frac{25}{36}$ (ii) Let E2 be the event in which 5 will come up on at least one.      Then the favourable outcomes are:     (1,5), (2,5), (3,5), (4,5), (5,5), ​(6,5), ​(5,1), ​(5,2), ​​(5,3), ​​(5,4) and ​​(5,6)     Number of favourable outcomes = 11 ∴ P ( 5 will come up on at least one) = P (E2) = $\frac{11}{36}$ (iii) Let E3 be the event in which 5 will come up on both the dice.      Then the favourable outcome is (5,5).     Number of favourable outcomes = 1 ∴ P (5 will come up on both the dice​) = P (E2) = $\frac{1}{36}$MathematicsRS Aggarwal (2015)Standard X

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