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 the probability that the event E occurs at least 3 times, is

• A. \( \left( \frac{24}{25} \right)^4 \)
• B. \( \left( \frac{4}{25} \right)^4 \)
• C. \( \left( \frac{7}{25} \right)^4 \)
• D. \( \left( \frac{96}{25} \right)^7 \)

Hint:

For binomial probability, use \( P(k \text{ successes}) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \), where \( p \) is the probability of success and \( n \) is the number of trials.

Answer 1

The Correct Option is A

Step 1: Identify the numbers where the product of the digits is 24.
The event E occurs if the product of the digits of a number is 24. For a two-digit number, let's analyze the possible pairs of digits whose product equals 24. These pairs are:
- \( (1, 24) \) (invalid since 24 is not a valid digit),
- \( (2, 12) \) (invalid),
- \( (3, 8) \), and
- \( (4, 6) \).
Thus, the valid numbers are 38, 84. These are the only numbers where the product of the digits equals 24.

Step 2: Probability of event E occurring for each number.
The total number of possible two-digit numbers (from 00 to 99) is 100. The two valid numbers are 38 and 84. Therefore, the probability of event E occurring (i.e., selecting a number whose digits multiply to 24) is:
\[ P(E) = \frac{2}{100} = \frac{1}{50}. \]

Step 3: Define the probability of E occurring at least 3 times.
We are selecting 4 numbers, so the probability that event E occurs at least 3 times is a binomial probability problem. We use the binomial probability formula:
\[ P(k \text{ successes}) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \]
where \( n = 4 \) (the number of trials), \( k \) is the number of successful outcomes (in this case, 3 or 4), and \( p = \frac{1}{50} \) is the probability of success. We want the probability of getting event E at least 3 times (i.e., 3 times or 4 times), so we compute:
\[ P(\text{at least 3 successes}) = P(3) + P(4) \]
where:
\[ P(3) = \binom{4}{3} \cdot \left(\frac{1}{50}\right)^3 \cdot \left(\frac{49}{50}\right)^1 \]
and
\[ P(4) = \binom{4}{4} \cdot \left(\frac{1}{50}\right)^4 \cdot \left(\frac{49}{50}\right)^0 \]

Step 4: Calculate the binomial coefficients and the probabilities.
Using the binomial coefficient:
\[ \binom{4}{3} = 4, \quad \binom{4}{4} = 1 \]
we calculate:
\[ P(3) = 4 \cdot \left(\frac{1}{50}\right)^3 \cdot \left(\frac{49}{50}\right) \]
and
\[ P(4) = \left(\frac{1}{50}\right)^4 \]
Summing these gives us the total probability of event E occurring at least 3 times.

Step 5: Conclusion.
Thus, the correct answer is:
\[ \boxed{\left( \frac{24}{25} \right)^4} \]