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(D) When two coins are tossed,the number of heads in one trial can be $0, 1,$ or $2$. In $5$ trials,the total number of tosses is $10$. Let $X$ be the

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$1/2$

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(D) When two coins are tossed,the number of heads in one trial can be $0, 1,$ or $2$. In $5$ trials,the total number of tosses is $10$. Let $X$ be the total number of heads in $10$ independent tosses of a single coin. $X$ follows a binomial distribution $B(n, p)$ with $n = 10$ and $p = 1/2$. We want the probability that $X$ is odd,i.e.,$P(X \in \{1, 3, 5, 7, 9\})$. $P(X 	ext{ is odd}) = \sum_{k \in \{1, 3, 5, 7, 9\}} \binom{10}{k} (\frac{1}{2})^k (\frac{1}{2})^{10-k} = (\frac{1}{2})^{10} \sum_{k \in \{1, 3, 5, 7, 9\}} \binom{10}{k}$. Using the property of binomial coefficients,$\sum_{k 	ext{ odd}} \binom{n}{k} = 2^{n-1}$. Here,$n = 10$,so the sum is $2^{10-1} = 2^9$. Therefore,$P(X 	ext{ is odd}) = \frac{2^9}{2^{10}} = \frac{1}{2}$.

Two coins are tossed $5$ times. What is the probability of getting an odd number of heads?

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