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Question
A pair of dice is thrown 4 times. If getting a doublet is considered as success, find the probability of two successes.
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Solution
The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.
Probability of getting doublets in a single throw of the pair of dice is
`p = 6/36`
= `1/6`
∴ q = 1 – p
= `1 - 1/6`
= `5/6`
Clearly, X has the binomial distribution with n = 4, `p = 1/6` and `q = 5/6`
∴ `P(X = x) = ""^nC_x q^(n - x)p^x`, where x = 0, 1, 2, 3 ... n
= `""^4C_x (5/6)^(4 - x) * (1/6)^x`
= `""^4C_x * 5^(4 - x)/6^4`
∴ P(2 successes) = P(X = 2)
= `""^4C_2 * 5^(4 - 2)/6^4`
= `6 * (25)/(1296)`
= `(25)/(216)`
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