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15 coins are tossed, then the probability of getting 10 heads will be
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Verified Answer
The correct answer is:
$\frac{3003}{32768}$
$\therefore$ Required probability $={ }^{15} C_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{5}$
$={ }^{15} C_{5} \frac{1}{2^{15}}$
$=\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{1}{2^{15}}$
$=\frac{3003}{32768}$
$={ }^{15} C_{5} \frac{1}{2^{15}}$
$=\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{1}{2^{15}}$
$=\frac{3003}{32768}$
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