Five coins,each having faces marked 2 and 3,a | vedclass

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(D) Let $X_i$ be the value obtained on the $i$-th coin,where $X_i \in \{2, 3\}$.We toss $5$ coins,so we have $X_1 + X_2 + X_3 + X_4 + X_5 = 12$.Let $n

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$\frac{5}{16}$

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(D) Let $X_i$ be the value obtained on the $i$-th coin,where $X_i \in \{2, 3\}$.We toss $5$ coins,so we have $X_1 + X_2 + X_3 + X_4 + X_5 = 12$.Let $n_2$ be the number of coins showing $2$ and $n_3$ be the number of coins showing $3$.Then $n_2 + n_3 = 5$ and $2n_2 + 3n_3 = 12$.Substituting $n_2 = 5 - n_3$ into the second equation: $2(5 - n_3) + 3n_3 = 12 \implies 10 - 2n_3 + 3n_3 = 12 \implies n_3 = 2$.Thus,$n_2 = 3$.This means we need exactly $3$ coins to show $2$ and $2$ coins to show $3$.Since each coin is fair,the probability of getting $2$ is $p = \frac{1}{2}$ and the probability of getting $3$ is $q = \frac{1}{2}$.Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$,where $n=5$ and $k=2$ (for the number of $3$s):$P = \binom{5}{2} \left(\frac{1}{2}ight)^2 \left(\frac{1}{2}ight)^3 = 10 	imes \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$.

Five coins,each having faces marked $2$ and $3$,are tossed. The probability of obtaining a total sum of $12$ is:

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