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Question
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than 4 appears on at least one die.
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Solution 1
When a die is tossed twice, the sample space S has 6 × 6 = 36 sample points.
∴ n(S) = 36
Trial will be a success if the number on at least one die is 5 or 6.
Let X denote the number of dice on which 5 or 6 appears.
Then X can take values 0, 1, 2
When X = 0 i.e., 5 or 6 do not appear on any of the dice, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}.
∴ n(X) = 16
∴ P(X = 0) =`(n(X))/(n(S)) = 16/36 = 4/9`
When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then
X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ n(X) = 16
∴ P(X = 1) = `(n(X))/(n(S)) = 16/36 = 4/9`
When X = 2, i.e. 5 or 6 appear on both of the dice, then
X = {(5, 5), (5, 6), (6, 5), (6, 6)}
∴ n(X) = 4
∴ P(X = 2) =`(n(X))/(n(S)) = 4/36 = 1/9`
∴ The required probability distribution is
| X | 0 | 1 | 2 |
| P(X = x) | `4/9` | `4/9` | `1/9` |
Solution 2
Success is defined as a number greater than 4 appears on at least one die
Let X denote the number of successes.
∴ Possible values of X and 0, 1, 2.
Let P(getting a number greater than 4) = p
= `2/6`
= `1/3`
∴ q = 1 – p
= `1 - 1/3`
= `2/3`
∴ P(X = 0) = P(no success)
= q2
= `4/9`
P(X = 1) = P(one success)
= qp + pq = 2pq
= `2 xx 1/3 xx 2/3`
= `4/9`
P(X = 2) = P(two successes)
= pp
= p2
= `1/9`
∴ Probability distribution of X is as follows:
| X | 0 | 1 | 2 |
| P(X = x) | `4/9` | `4/9` | `1/9` |
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