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 draw $2$ specific tickets ($15$ and $89$) and $3$ other tickets from t

Vedclass · Accepted Answer

$\frac{2}{801}$

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(A) Total number of ways to draw $5$ tickets from $90$ is $^{90}C_5$.We need to draw $2$ specific tickets ($15$ and $89$) and $3$ other tickets from the remaining $88$ tickets.The number of favorable outcomes is $^{2}C_2 	imes ^{88}C_3 = 1 	imes ^{88}C_3$.The probability $P = \frac{^{88}C_3}{^{90}C_5}$.$P = \frac{88 	imes 87 	imes 86}{3 	imes 2 	imes 1} \div \frac{90 	imes 89 	imes 88 	imes 87 	imes 86}{5 	imes 4 	imes 3 	imes 2 	imes 1}$.$P = \frac{88 	imes 87 	imes 86}{6} 	imes \frac{120}{90 	imes 89 	imes 88 	imes 87 	imes 86}$.$P = \frac{120}{90 	imes 89 	imes 6} = \frac{20}{89 	imes 6} = \frac{10}{89 	imes 3} = \frac{10}{267}$.Wait,re-calculating: $P = \frac{88! / (3! 85!)}{90! / (5! 85!)} = \frac{88!}{3! 85!} 	imes \frac{5! 85!}{90!} = \frac{5 	imes 4}{90 	imes 89} = \frac{20}{8010} = \frac{2}{801}$.

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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