Five numbers x_{1}, x_{2}, x_{3}, x_{4}, x_{5 | vedclass

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(B) The total number of ways to select $5$ distinct numbers from $18$ is given by $n(S) = {}^{18}C_{5} = \frac{18 	imes 17 	imes 16 	imes 15 	imes

Vedclass · Accepted Answer

$\frac{1}{68}$

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(B) The total number of ways to select $5$ distinct numbers from $18$ is given by $n(S) = {}^{18}C_{5} = \frac{18 	imes 17 	imes 16 	imes 15 	imes 14}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 8568$. For the condition $x_{2} = 7$ and $x_{4} = 11$ to hold,we must have: $x_{1} $x_{3}$ must be between $7$ and $11$ (where $x_{3} \in \{8, 9, 10\}$),so there are ${}^{3}C_{1} = 3$ ways. $x_{5} > 11$ (where $x_{5} \in \{12, 13, 14, 15, 16, 17, 18\}$),so there are ${}^{7}C_{1} = 7$ ways. The number of favorable outcomes is $n(E) = 6 	imes 3 	imes 7 = 126$. The probability $P(E) = \frac{n(E)}{n(S)} = \frac{126}{8568} = \frac{1}{68}$.

Five numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are randomly selected from the numbers $1, 2, 3, \ldots, 18$ and are arranged in increasing order $(x_{1} < x_{2} < x_{3} < x_{4} < x_{5})$. The probability that $x_{2} = 7$ and $x_{4} = 11$ is

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