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प्रश्न
Find the probability distribution of the number of doublets in three throws of a pair of dice
उत्तर
Let X denotes the number of doublets in three throws of a pair of dice.
∴ Possible values of X are 0, 1, 2 and 3.
In a toss of pair of dice, possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).
∴ Probability of getting a doublet = p
= `6/36`
= `1/6`
∴ Probability of not getting a doublet = q
= `1 - 1/6`
= `5/6`
∴ P(X = 0) = P(no doublet)
= q × q × q
= `5/6 xx 5/6 xx 5/6`
= `125/216`
P(X = 1) = P(one doublet)
= pqq + qpq + qqp
= `1/6 xx 5/6 xx 5/6 + 5/6 xx 1/6 xx 5/6 + 5/6 xx 5/6 xx 1/6`
= `75/216`
P(X = 2) = P(two doublets)
= ppq + pqp + qpp
= `1/6 xx 1/6 xx 5/6 + 1/6 xx 5/6 xx 1/6 + 5/6 xx 1/6 xx 1/6`
= `15/216`
P(X = 3) = P(three doublets)
= p × p × p
= `1/6 xx 1/6 xx 1/6`
= `1/216`
∴ Probability distribution of X is as follows:
| X | 0 | 1 | 2 | 3 |
| P(X = x) | `125/216` | `75/216` | `15/216` | `1/216` |
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