A box contains 50 tickets numbered 1, 2, 3, d | vedclass

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Question

(A) The total number of ways to choose $5$ tickets out of $50$ is $^{50}C_5$.For $x_3 = 30$,we must choose two numbers from the set $\{1, 2, \dots, 29

Vedclass · Accepted Answer

$\frac{^{29}C_2 	imes ^{20}C_2}{^{50}C_5}$

Vedclass · Answer

(A) The total number of ways to choose $5$ tickets out of $50$ is $^{50}C_5$.For $x_3 = 30$,we must choose two numbers from the set $\{1, 2, \dots, 29\}$ to be $x_1$ and $x_2$,and two numbers from the set $\{31, 32, \dots, 50\}$ to be $x_4$ and $x_5$.The number of ways to choose $x_1$ and $x_2$ is $^{29}C_2$.The number of ways to choose $x_4$ and $x_5$ is $^{50-30}C_2 = ^{20}C_2$.Thus,the number of favorable outcomes is $^{29}C_2 	imes ^{20}C_2$.The probability is $\frac{^{29}C_2 	imes ^{20}C_2}{^{50}C_5}$.

$A$ box contains $50$ tickets numbered $1, 2, 3, \dots, 50$. Five tickets are drawn at random and arranged in ascending order $(x_1 < x_2 < x_3 < x_4 < x_5)$. What is the probability that $x_3 = 30$?

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