Question

Numbers are selected at random, one at a time from the two-digit numbers \(00,01,02,-------, 99\) 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 probability, that the event E occurs at least 3 times, is

• A. \(\frac{24}{(25)^{4}}\)
• B. \(\frac{4}{(25)^{4}}\)
• C. \(\frac{97}{(25)^{4}}\)
• D. \(\frac{96}{(25)^{4}}\)

Hint:

Find the chance that one selected number has digit product 24, then apply the binomial formula for at least 3 successes in 4 trials.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Each draw is independent because of replacement. We first find the probability \(p\) that a two-digit number \(00\) to \(99\) has digit product 24.

Step 2: Find p
Digit pairs with product 24 are \((3,8),(8,3),(4,6),(6,4)\), that is 4 numbers out of 100.
So \(p=\frac{4}{100}=\frac{1}{25}\) and \(q=\frac{24}{25}\).

Step 3: Binomial calculation
\(P(X\geq3)=P(3)+P(4)=\binom{4}{3}\left(\frac{1}{25}\right)^3\frac{24}{25}+\left(\frac{1}{25}\right)^4\)
\(=\frac{4\cdot24+1}{25^4}=\frac{97}{25^4}\).

Step 4: Compare the options
This matches option (C).

Final Answer:
The required probability is \(\frac{97}{(25)^4}\). \[ \boxed{\text{(C) }\dfrac{97}{(25)^{4}}} \]