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(C) The total number of two-digit numbers from $00$ to $99$ is $100$. Let $X$ be the number formed by digits $d_1 d_2$. The product $d_1 	imes d_2 =

Vedclass · Accepted Answer

$\frac{97}{(25)^4}$

Vedclass · Answer

(C) The total number of two-digit numbers from $00$ to $99$ is $100$. Let $X$ be the number formed by digits $d_1 d_2$. The product $d_1 	imes d_2 = 24$. The possible pairs $(d_1, d_2)$ are $(3, 8), (4, 6), (6, 4), (8, 3)$. Thus,there are $4$ such numbers: $38, 46, 64, 83$. The probability of event $E$ occurring in a single trial is $p = \frac{4}{100} = \frac{1}{25}$. The probability of event $E$ not occurring is $q = 1 - p = 1 - \frac{1}{25} = \frac{24}{25}$. We select $n = 4$ numbers. Let $X$ be the number of times event $E$ occurs. $X$ follows a binomial distribution $B(n, p) = B(4, \frac{1}{25})$. We need to find $P(X \ge 3) = P(X = 3) + P(X = 4)$. $P(X = 3) = \binom{4}{3} p^3 q^1 = 4 	imes (\frac{1}{25})^3 	imes (\frac{24}{25}) = \frac{4 	imes 24}{(25)^4} = \frac{96}{(25)^4}$. $P(X = 4) = \binom{4}{4} p^4 q^0 = 1 	imes (\frac{1}{25})^4 	imes 1 = \frac{1}{(25)^4}$. $P(X \ge 3) = \frac{96}{(25)^4} + \frac{1}{(25)^4} = \frac{97}{(25)^4}$.

Numbers are selected at random,one at a time from the two-digit numbers $00, 01, 02, \dots, 99$ with replacement. An event $E$ occurs only if the product of the two digits of a selected number is $24$. If four numbers are selected,then the probability that the event $E$ occurs at least $3$ times is:

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