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(C) Let $X$ be the number of heads obtained by person $A$ and $Y$ be the number of heads obtained by person $B$. Both $X$ and $Y$ follow a binomial di

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$\frac{5}{16}$

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(C) Let $X$ be the number of heads obtained by person $A$ and $Y$ be the number of heads obtained by person $B$. Both $X$ and $Y$ follow a binomial distribution $B(n=3, p=1/2)$.The probability of getting $k$ heads in $3$ tosses is given by $P(X=k) = \binom{3}{k} (1/2)^3 = \binom{3}{k} / 8$.We want to find $P(X=Y) = P(X=0, Y=0) + P(X=1, Y=1) + P(X=2, Y=2) + P(X=3, Y=3)$.Since $A$ and $B$ are independent,$P(X=k, Y=k) = P(X=k) 	imes P(Y=k) = [P(X=k)]^2$.$P(X=0) = \binom{3}{0}/8 = 1/8 \implies P(X=0, Y=0) = (1/8)^2 = 1/64$.$P(X=1) = \binom{3}{1}/8 = 3/8 \implies P(X=1, Y=1) = (3/8)^2 = 9/64$.$P(X=2) = \binom{3}{2}/8 = 3/8 \implies P(X=2, Y=2) = (3/8)^2 = 9/64$.$P(X=3) = \binom{3}{3}/8 = 1/8 \implies P(X=3, Y=3) = (1/8)^2 = 1/64$.Total probability $= 1/64 + 9/64 + 9/64 + 1/64 = 20/64 = 5/16$.

Each of the persons $A$ and $B$ independently tosses three fair coins. The probability that both of them get the same number of heads is:

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