Question

Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws. 

Answer 1

\[\text{ Let getting a multiple of 3 be a success}  . \]
\[\text{ We have } , \]
\[p = \text{ the probability of getting a success } = \frac{2}{6} = \frac{1}{3}\]
\[\text{ So, q = the probability of getting a failure } = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}\]
\[\text{ Let X denote the number of success in a sample of 10 trials . Then } , \]
\[X \text{  follows binomial distribution with parameters } n = 10, p = \frac{1}{3} \text{ and } q = \frac{2}{3}\]
\[ \therefore P\left( X = r \right) = ^{10}{}{C}_r p^r q^\left( 10 - r \right) = ^{10}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^\left( 10 - r \right) = ^{10}{}{C}_r \left( \frac{1}{3} \right)^{10} 2^\left( 10 - r \right) = \frac{^{10}{}{C}_r 2^\left( 10 - r \right)}{3^{10}}, r = 0, 1, 2, . . . , 10\]
\[\text{ Now } , \]
\[\text{ Required probability}  = P\left( X \geq 8 \right)\]
\[ = P\left( X = 8 \right) + P\left( X = 9 \right) + P\left( X = 10 \right)\]
\[ = \frac{^{10}{}{C}_8 2^\left( 10 - 8 \right)}{3^{10}} + \frac{^{10}{}{C}_9 2^\left( 10 - 9 \right)}{3^{10}} + \frac{^{10}{}{C}_{10} 2^\left( 10 - 10 \right)}{3^{10}}\]
\[ = \frac{45 \times 2^2}{3^{10}} + \frac{10 \times 2}{3^{10}} + \frac{1}{3^{10}}\]
\[ = \frac{180 + 20 + 1}{3^{10}}\]
\[ = \frac{201}{3^{10}}\]