If getting a head on a coin when it is tossed | vedclass

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(C) Let $n=10$ be the number of tosses, $p$ be the probability of success (getting a head) $= \frac{1}{2}$, and $q$ be the probability of failure (get

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$\frac{193}{2^9}$

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(C) Let $n=10$ be the number of tosses, $p$ be the probability of success (getting a head) $= \frac{1}{2}$, and $q$ be the probability of failure (getting a tail) $= \frac{1}{2}$.Let $X$ be the number of successes. We want the probability of having more failures than successes. This means $X However, the question asks for more failures than successes, which means $X  X$, which implies $2X Thus, we need to calculate $P(X \leq 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$.Since the distribution is symmetric for $p=q=\frac{1}{2}$, $P(X \leq 4) = P(X \geq 6)$.$P(X \geq 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)$.$P(X \geq 6) = \sum_{r=6}^{10} {}^{10}C_r (\frac{1}{2})^{10} = \frac{1}{2^{10}} ({}^{10}C_6 + {}^{10}C_7 + {}^{10}C_8 + {}^{10}C_9 + {}^{10}C_{10})$.$= \frac{1}{2^{10}} (210 + 120 + 45 + 10 + 1) = \frac{386}{2^{10}} = \frac{193}{2^9}$.

If getting a head on a coin when it is tossed is considered as success, then the probability of having more number of failures when ten fair coins are tossed simultaneously is

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