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A coin is tossed 7 times. Each time a man calls head. Find the probability that he wins the toss on more occasions.
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1401 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
The man has to win at least 4 times.
$$
\begin{array}{r}
\therefore \text { Reqd. probability }= \\
\qquad \begin{array}{r}
7 \mathrm{C}_{4}\left(\frac{1}{2}\right)^{4} \cdot\left(\frac{1}{2}\right)^{3}+{ }^{7} \mathrm{C}_{5}\left(\frac{1}{2}\right)^{5} \cdot\left(\frac{1}{2}\right)^{2} \\
+{ }^{7} \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6} \cdot \frac{1}{2}+{ }^{7} \mathrm{C}_{7}\left(\frac{1}{2}\right)^{7} \\
=\left({ }^{7} \mathrm{C}_{4}+{ }^{7} \mathrm{C}_{5}+{ }^{7} \mathrm{C}_{6}+{ }^{7} \mathrm{C}_{7}\right) \cdot \frac{1}{2^{7}}=\frac{64}{2^{7}}=\frac{1}{2}
\end{array}
\end{array}
$$
$$
\begin{array}{r}
\therefore \text { Reqd. probability }= \\
\qquad \begin{array}{r}
7 \mathrm{C}_{4}\left(\frac{1}{2}\right)^{4} \cdot\left(\frac{1}{2}\right)^{3}+{ }^{7} \mathrm{C}_{5}\left(\frac{1}{2}\right)^{5} \cdot\left(\frac{1}{2}\right)^{2} \\
+{ }^{7} \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6} \cdot \frac{1}{2}+{ }^{7} \mathrm{C}_{7}\left(\frac{1}{2}\right)^{7} \\
=\left({ }^{7} \mathrm{C}_{4}+{ }^{7} \mathrm{C}_{5}+{ }^{7} \mathrm{C}_{6}+{ }^{7} \mathrm{C}_{7}\right) \cdot \frac{1}{2^{7}}=\frac{64}{2^{7}}=\frac{1}{2}
\end{array}
\end{array}
$$
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