Question

Six coins are tossed simultaneously. Find the probability of getting
(i) 3 heads
(ii) no heads
(iii) at least one head

Answer 1

Let X denote the number of heads obtained in tossing 6 coins.
Then, X follows a binomial distribution with n=6,

\[p = \frac{1}{2} \text{ and }  q = \frac{1}{2}\]
\[\text{ Hence, the distribution is given by } \]
\[P(X = r) = ^{6}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{6 - r} , r = 0, 1, 2, 3, 4, 5, 6\]
\[ = \frac{^{6}{}{C}_r}{2^6}\]
\[\left( i \right) P(\text{ getting  3 heads }) = P(X = 3)\]
\[ = \frac{^{6}{}{C}_3}{2^6}\]
\[ = \frac{20}{64}\]
\[ = \frac{5}{16}\]
\[\left( ii \right) P(\text{ getting no head } ) = P(X = 0) \]
\[ = \frac{^{6}{}{C}_0}{2^6}\]
\[ = \left( \frac{1}{2} \right)^6 \]
\[ = \frac{1}{64}\]
\[\left( iii \right) P(\text{ getting at least 1 head} ) = P(X \geq 1) \]
\[ = 1 - P(X = 0) \]
\[ = 1 - \frac{1}{64}\]
\[ = \frac{63}{64}\]