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(C) The total number of two-digit numbers from $10$ to $99$ is $99 - 10 + 1 = 90$. Event $E$ occurs if the product of the two digits is $18$. The poss

Vedclass · Accepted Answer

$87\left(\frac{4}{90}\right)^4$

Vedclass · Answer

(C) The total number of two-digit numbers from $10$ to $99$ is $99 - 10 + 1 = 90$. Event $E$ occurs if the product of the two digits is $18$. The possible numbers are $\{29, 36, 63, 92\}$. Thus,the probability of event $E$ occurring in a single trial is $p = \frac{4}{90}$. The probability of event $E$ not occurring is $q = 1 - p = 1 - \frac{4}{90} = \frac{86}{90}$. Since $n = 4$ numbers are selected with replacement,we use the binomial distribution $P(X = k) = \binom{n}{k} p^k q^{n-k}$. We need to find the probability that event $E$ occurs at least $3$ times,i.e.,$P(X \geq 3) = P(X = 3) + P(X = 4)$. $P(X = 3) = \binom{4}{3} \left(\frac{4}{90}ight)^3 \left(\frac{86}{90}ight)^1 = 4 	imes \frac{4^3}{90^3} 	imes \frac{86}{90} = \frac{4 	imes 64 	imes 86}{90^4} = \frac{22016}{90^4}$. $P(X = 4) = \binom{4}{4} \left(\frac{4}{90}ight)^4 \left(\frac{86}{90}ight)^0 = 1 	imes \frac{4^4}{90^4} 	imes 1 = \frac{256}{90^4}$. $P(X \geq 3) = \frac{22016 + 256}{90^4} = \frac{22272}{90^4}$. Simplifying the expression: $P(X \geq 3) = 4 \left(\frac{4}{90}ight)^3 \left(\frac{86}{90}ight) + \left(\frac{4}{90}ight)^4 = \left(\frac{4}{90}ight)^3 \left(4 	imes \frac{86}{90} + \frac{4}{90}ight) = \left(\frac{4}{90}ight)^3 \left(\frac{344 + 4}{90}ight) = \left(\frac{4}{90}ight)^3 \left(\frac{348}{90}ight) = \frac{348}{90} 	imes \left(\frac{4}{90}ight)^3 = 87 	imes \frac{4}{90} 	imes \left(\frac{4}{90}ight)^3 = 87 \left(\frac{4}{90}ight)^4$.

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