A pair of dice is thrown 6 times. If getting a total of 7 is considered a success, find the probability of at least five successes.

Solution

Here n = 6
In a throw of pair dice, we get total of 7 when get
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
$therefore space space space straight p space equals space 6 over 36 space equals space 1 over 6 comma space space straight q space equals space 1 space minus space straight p space equals space 1 space minus space 1 over 6 space equals space 5 over 6$
P(at least 5 successes) = P(5) + P(6)
$equals space straight C presuperscript 6 subscript 5 space open parentheses 5 over 6 close parentheses space open parentheses 1 over 6 close parentheses to the power of 5 space plus space straight C presuperscript 6 subscript 6 space open parentheses 1 over 6 close parentheses to the power of 6 space equals space 6 over 1 cross times 5 over 6 cross times 1 over 6 to the power of 5 plus 1 space cross times 1 over 6 to the power of 6 space equals 1 over 6 to the power of 6 left parenthesis 30 plus 1 right parenthesis equals space fraction numerator 31 over denominator 6 to the power of 6 space end fraction space equals 31 over 130896$

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Probability

A pair of dice is thrown 6 times. If getting a total of 7 is considered a success, find the probability of at least five successes.

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