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: Understanding the Question:
We are selecting four numbers from the set {00, 01, ..., 99}. The event E occurs if the product of the digits of a number is 24. We need the probability that E occurs 3 or 4 times.
Step 2: Key Formula or Approach:
This follows a Binomial Distribution \( \text{X} \sim \text{B}(n, p) \).
Formula: \( \text{P(X} = r) = \binom{n}{r} p^r q^{n-r} \).
Step 3: Detailed Explanation:
1. Total numbers = 100.
2. Favorable numbers for product = 24: {38, 46, 64, 83}.
Number of favorable outcomes = 4.
3. Probability of success, \( p = \frac{4}{100} = \frac{1}{25} \).
Probability of failure, \( q = 1 - p = \frac{24}{25} \).
4. Number of trials, \( n = 4 \). We want \( \text{P(X} \ge 3) \).
\[ \text{P(X} \ge 3) = \text{P(X} = 3) + \text{P(X} = 4) \] \[ \text{P(X} = 3) = \binom{4}{3} \left( \frac{1}{25} \right)^3 \left( \frac{24}{25} \right)^1 = 4 \times \frac{24}{25^4} = \frac{96}{25^4} \] \[ \text{P(X} = 4) = \binom{4}{4} \left( \frac{1}{25} \right)^4 \left( \frac{24}{25} \right)^0 = 1 \times \frac{1}{25^4} = \frac{1}{25^4} \] \[ \text{Total Probability} = \frac{96 + 1}{25^4} = \frac{97}{25^4} \] Step 4: Final Answer:
The probability is \( \frac{97}{(25)^4} \).