Seven chits are numbered 1 to 7. Three are dr | vedclass

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(C) Let $X_1, X_2, X_3$ be the numbers on the three chits drawn with replacement. Each $X_i \in \{1, 2, 3, 4, 5, 6, 7\}$.We want the probability that

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$(\frac{3}{7})^3 - (\frac{2}{7})^3$

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(C) Let $X_1, X_2, X_3$ be the numbers on the three chits drawn with replacement. Each $X_i \in \{1, 2, 3, 4, 5, 6, 7\}$.We want the probability that $\min(X_1, X_2, X_3) = 5$.This is equivalent to the condition that all $X_i \ge 5$ $AND$ at least one $X_i = 5$.$P(\min \ge 5) = P(X_i \ge 5 	ext{ for all } i) = (\frac{3}{7})^3$.$P(\min \ge 6) = P(X_i \ge 6 	ext{ for all } i) = (\frac{2}{7})^3$.$P(\min = 5) = P(\min \ge 5) - P(\min \ge 6) = (\frac{3}{7})^3 - (\frac{2}{7})^3$.

Seven chits are numbered $1$ to $7$. Three are drawn one by one with replacement. The probability that the least number on any selected chit is $5$ is:

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