Question

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times. 

Answer 1

Let getting an odd number be a success in a trial.
We have,

\[p = \text{ probability of getting an odd number in a trial } = \frac{3}{6} = \frac{1}{2}\]
\[\text{ Also } , q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}\]
\[\text{ Let X denote the number of success in a sample of 5 trials . Then,}  \]
\[\text{ X follows binomial distribution with parameters n = 5 and } p = q = \frac{1}{2}\]
\[ \therefore P\left( X = r \right) = ^{5}{}{C}_r p^r q^\left( 5 - r \right) = ^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^\left( 5 - r \right) = ^{5}{}{C}_r \left( \frac{1}{2} \right)^5 , \text{where}  r = 0, 1, 2, 3, 4, 5\]
\[\text{ Now,}  \]
\[\text{ Required probability}  = P\left( X = 3 \right)\]
\[ = ^{5}{}{C}_3 \left( \frac{1}{2} \right)^5 \]
\[ = \frac{10}{32}\]
\[ = \frac{5}{16}\]