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Question
A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?
Solution
Probability of success = Probability of getting a total of 9
= Probability of getting (3,6), (4,5), (5,4), (6,3) out of 36 outcomes
\[p = \frac{4}{36} = \frac{1}{9}, q = 1 - p = \frac{8}{9}\text{ and } n = 6\]
\[\text{ X follows a binomial distribution with n} = 6, p = \frac{1}{9} \text{ and } q = \frac{8}{9}\]
\[P(X = r) = ^{6}{}{C}_r \left( \frac{1}{9} \right)^r \left( \frac{8}{9} \right)^{6 - r} \]
\[\text{ The required probability = Probability of at least 5 successes } \]
\[ = P(X \geq 5) \]
\[ = P(X = 5) + P(X = 6)\]
\[ = ^{6}{}{C}_5 \left( \frac{1}{9} \right)^5 \left( \frac{8}{9} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{9} \right)^6 \left( \frac{8}{9} \right)^{6 - 6} \]
\[ = \frac{6(8) + 1}{9^6}\]
\[ = \frac{49}{9^6}\]
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