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Question
In a simultaneous throw of a pair of dice, find the probability of getting 1 at least once
Solution
In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) which 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)
E ⟶ event of getting a 1 at least once
No. of favorable outcomes = 11 {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 1) (3, 1)
(4, 1) (5, 1) (6, 1)}
Total no. of possible outcomes = 36
We know that, Probability P(E) =`"No.of favorable outcomes"/"Total no.of possible outcomes"`
P(E) =11/36
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