A pair of dice is thrown 4 times. If a doublet is considered a success, find the probability of 2 successes.

Solution

Here n = 4
Let p be the probability that doublet is obtained
$therefore space space straight p space equals space straight P open curly brackets left parenthesis 1 comma space 1 right parenthesis comma space left parenthesis 2 comma space 2 right parenthesis comma space left parenthesis 3 comma space 3 right parenthesis comma space left parenthesis 4 comma space 4 right parenthesis comma space left parenthesis 5 comma space 5 right parenthesis comma space left parenthesis 6 comma space 6 right parenthesis close curly brackets space equals space 6 over 36 equals 1 over 6 therefore space space space space straight q space equals space 1 space minus space straight p space equals space 1 space minus space 1 over 6 space equals 5 over 6 Required space probability space space equals space straight P left parenthesis 2 right parenthesis space equals space straight C presuperscript 4 subscript 2 space open parentheses 5 over 6 close parentheses squared space open parentheses 1 over 6 close parentheses squared space equals space fraction numerator 4 space cross times space 3 over denominator 1 cross times 2 end fraction cross times 25 over 36 space equals 25 over 216$

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Probability

A pair of dice is thrown 4 times. If a doublet is considered a success, find the probability of 2 successes.

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