Question

A coin is tossed 5 times. What is the probability of getting at least 3 heads?

 

Answer 1

\[\text{ Let X denote the number of heads in 5 tosses }  . \]
\[ \text{ X follows a binomial distribution with n } = 5; p = \text{ probability of getting a head } = \frac{1}{2}\text{ and }  q = 1 - p = \frac{1}{2}\]
\[P(X = r) =^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{5 - r} , r = 0, 1, 2 . . . 5\]
\[\text{ The required probability = P(getting at least 3 heads } )\]
\[ = P(X \geq 3) \]
\[ = P(X = 3) + P(X = 4) + P(X = 5)\]
\[ = ^{5}{}{C}_3 \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^{5 - 3} + ^{5}{}{C}_4 \left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^{5 - 1} + ^{5}{}{C}_5 \left( \frac{1}{2} \right)^5 \left( \frac{1}{2} \right)^{5 - 0} \]

\[= 10 \left( \frac{1}{2} \right)^5 + 5 \left( \frac{1}{2} \right)^5 + 1 \left( \frac{1}{2} \right)^5 \]

\[ = \left( \frac{1}{2} \right)^5 (10 + 5 + 1) \]

\[ = \left( \frac{1}{2} \right)^5 \times 16 \]

\[ = \frac{1}{2}\]