Question

Find the probability of getting a sum of 9 at least twice in 10 throws with two dice.

Solution

Here n = 10
We get total of 9 when we get, (3, 6), (4, 5), (5, 4), (6, 3)
$therefore space space space space straight p space equals space 4 over 36 space equals space 1 over 9 comma space space straight q space equals space 1 minus space straight p space equals space 1 space minus space 1 over 9 space equals space 8 over 9$
P(getting sum 9 at least twice) = 1 - [P(0) + P(1)]
$equals space 1 minus space open square brackets straight C presuperscript 10 subscript 0 space open parentheses 1 over 9 close parentheses to the power of 0 space open parentheses 8 over 9 close parentheses to the power of 10 space plus space straight C presuperscript 10 subscript 1 space open parentheses 1 over 9 close parentheses to the power of 1 space open parentheses 8 over 9 close parentheses to the power of 9 close square brackets space equals space 1 minus open square brackets 1 space cross times space 1 space cross times space open parentheses 8 over 9 close parentheses to the power of 10 plus space 10 over 1 cross times 1 over 9 cross times open parentheses 8 over 9 close parentheses to the power of 9 close square brackets space equals 1 minus open parentheses 8 over 9 close parentheses to the power of 9 space open square brackets 8 over 9 plus 10 over 9 close square brackets space equals space 1 minus open parentheses 8 over 9 close parentheses squared space left parenthesis 2 right parenthesis space equals space 1 minus space 2 space open parentheses 8 over 9 close parentheses to the power of 9$

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Probability

Find the probability of getting a sum of 9 at least twice in 10 throws with two dice.

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