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Question
An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 6 heads.
Solution
Let X be the number of heads in tossing the coin 8 times.
X follows a binomial distribution with n = 8
\[p = \frac{1}{2} \text{ and q } = \frac{1}{2}\]
\[\text{ Hence, the distribution is given by } \]
\[ \therefore P(X = r) = ^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r} , r = 0, 1, 2, 3, 4, 5, 6, 7, 8\]
\[\text{ Required probability } = P(X \geq 6)\]
\[ = P(X = 6) + P(X = 7) + P(X = 8)\]
\[ = \frac{^{8}{}{C}_6 + ^{8}{}{C}_7 + ^{8}{}{C}_8}{2^8}\]
\[ = \frac{28 + 8 + 1}{256}\]
\[ = \frac{37}{256}\]
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