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(A) Let $n = 100$ be the number of tosses and $p = q = \frac{1}{2}$ be the probability of getting a head or tail in a single toss.The probability of g

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$\frac{1}{2}$

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(A) Let $n = 100$ be the number of tosses and $p = q = \frac{1}{2}$ be the probability of getting a head or tail in a single toss.The probability of getting exactly $r$ heads is given by the binomial distribution: $P(X = r) = \binom{n}{r} p^r q^{n-r} = \binom{100}{r} (\frac{1}{2})^{100}$.We want the probability of getting an even number of heads,which is $S = \sum_{r 	ext{ is even}} \binom{100}{r} (\frac{1}{2})^{100}$.Using the binomial expansion,we know $(p+q)^n = \sum_{r=0}^{n} \binom{n}{r} p^r q^{n-r} = 1$ and $(q-p)^n = \sum_{r=0}^{n} \binom{n}{r} (-p)^r q^{n-r}$.For $p=q=\frac{1}{2}$,$(q-p)^n = 0^n = 0$ for $n \ge 1$.Thus,$\sum_{r=0}^{n} \binom{n}{r} (\frac{1}{2})^n = 1$ and $\sum_{r=0}^{n} \binom{n}{r} (-1)^r (\frac{1}{2})^n = 0$.Adding these two equations: $2 \sum_{r 	ext{ is even}} \binom{n}{r} (\frac{1}{2})^n = 1 + 0 = 1$.Therefore,the probability of an even number of heads is $S = \frac{1}{2}$.

$A$ fair coin is tossed $100$ times. The chance of getting a head an even number of times is

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