In a lottery,there were 90 tickets numbered 1 | vedclass

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(A) Total number of ways to draw $5$ tickets from $90$ is $^{90}C_5$. We need to select $2$ specific tickets ($15$ and $89$) and $3$ other tickets fro

Vedclass · Accepted Answer

$\frac{2}{801}$

Vedclass · Answer

(A) Total number of ways to draw $5$ tickets from $90$ is $^{90}C_5$. We need to select $2$ specific tickets ($15$ and $89$) and $3$ other tickets from the remaining $88$ tickets. The number of favorable outcomes is $^{2}C_2 	imes ^{88}C_3$. The probability $P = \frac{^{2}C_2 	imes ^{88}C_3}{^{90}C_5} = \frac{1 	imes \frac{88 	imes 87 	imes 86}{3 	imes 2 	imes 1}}{\frac{90 	imes 89 	imes 88 	imes 87 	imes 86}{5 	imes 4 	imes 3 	imes 2 	imes 1}} = \frac{88 	imes 87 	imes 86}{6} 	imes \frac{120}{90 	imes 89 	imes 88 	imes 87 	imes 86} = \frac{120}{90 	imes 89 	imes 6} = \frac{20}{90 	imes 89} = \frac{2}{9 	imes 89} = \frac{2}{801}$.

In a lottery,there were $90$ tickets numbered $1$ to $90$. Five tickets were drawn at random. What is the probability that two of the tickets drawn are numbered $15$ and $89$?

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