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?

Answer 1

Let X be the number of successes in 6 throws of the two dice.
 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

\[p = \frac{4}{36} = \frac{1}{9}, q = 1 - p = \frac{8}{9}\text{ and }  n = 6\]

\[\text{ X follows a binomial distribution with n}  = 6, p = \frac{1}{9} \text{ and } q = \frac{8}{9}\]

\[P(X = r) = ^{6}{}{C}_r \left( \frac{1}{9} \right)^r \left( \frac{8}{9} \right)^{6 - r} \]

\[\text{ The required probability = Probability of at least 5 successes } \]

\[ = P(X \geq 5) \]

\[ = P(X = 5) + P(X = 6)\]

\[ = ^{6}{}{C}_5 \left( \frac{1}{9} \right)^5 \left( \frac{8}{9} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{9} \right)^6 \left( \frac{8}{9} \right)^{6 - 6} \]

\[ = \frac{6(8) + 1}{9^6}\]

\[ = \frac{49}{9^6}\]