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Question
Eight coins are thrown simultaneously. Find the chance of obtaining at least six heads.
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Solution
Let X be the number of heads in tossing 8 coins.
X follows a binomial distribution with n =8;
\[p = \frac{1}{2}\text{ and } q = \frac{1}{2};\]
\[P(X = r) = ^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r} = ^{8}{}{C}_r \left( \frac{1}{2} \right)^8 \]
\[\text{ Probability of obtaining at least 6 heads} = P(X \geq 6) \]
\[ = P(X = 6) + P(X = 7) + P(X = 8)\]
\[ =^{8}{}{C}_6 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_7 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_8 \left( \frac{1}{2} \right)^8 \]
\[ = \frac{1}{2^8}\left( 28 + 8 + 1 \right) \]
\[ = \frac{37}{256}\]
\[\text{ Probability of obtaining at least 6 heads} = P(X \geq 6) \]
\[ = P(X = 6) + P(X = 7) + P(X = 8)\]
\[ =^{8}{}{C}_6 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_7 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_8 \left( \frac{1}{2} \right)^8 \]
\[ = \frac{1}{2^8}\left( 28 + 8 + 1 \right) \]
\[ = \frac{37}{256}\]
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