A bag contains books numbered 1 to 20. Three | vedclass

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Question

(D) Let $X_1, X_2, X_3$ be the numbers on the three books drawn with replacement. Each $X_i \in \{1, 2, \dots, 20\}$.Total number of outcomes $= 20^3$

Vedclass · Accepted Answer

$\left(\frac{7}{20}\right)^3-\left(\frac{6}{20}\right)^3$

Vedclass · Answer

(D) Let $X_1, X_2, X_3$ be the numbers on the three books drawn with replacement. Each $X_i \in \{1, 2, \dots, 20\}$.Total number of outcomes $= 20^3$.We want the largest number to be $7$. This means all three numbers must be $\le 7$,and at least one number must be $7$.Let $E$ be the event that the largest number is $\le 7$. Then $P(E) = \left(\frac{7}{20}\right)^3$.Let $F$ be the event that the largest number is $\le 6$. Then $P(F) = \left(\frac{6}{20}\right)^3$.The probability that the largest number is exactly $7$ is $P(E) - P(F) = \left(\frac{7}{20}\right)^3 - \left(\frac{6}{20}\right)^3$.Therefore,option $D$ is correct.

$A$ bag contains books numbered $1$ to $20$. Three books are drawn from the bag with replacement. The probability that the largest number on the books is $7$ is:

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