Numberse are selected at random, one at a time, from the two-digit numbers 00,01,02,….99 with replacement. An event E occurs if and only if the product of the two digits of a selected number is 18. If four numbers are selected, find probability that the event E occurs at least 3 times.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Identify the event E
The event E occurs if the product of the two digits of a selected number is 18. We need to find out which two-digit numbers satisfy this condition.

The two-digit numbers can be represented as \( ab \) where \( a \) is the tens digit and \( b \) is the units digit. The product \( a \times b = 18 \).

The pairs \( (a, b) \) that satisfy this are:
- \( (2, 9) \) → Number 29
- \( (3, 6) \) → Number 36
- \( (6, 3) \) → Number 63
- \( (9, 0) \) → Number 90

Thus, the valid numbers are: 29, 36, 63, and 90.

### Step 2: Calculate the probability of event E
The total number of two-digit numbers from 00 to 99 is 100. The favorable outcomes for event E are 4 (the numbers identified in Step 1).

The probability \( P \) of event E occurring is given by:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{100} = \frac{1}{25} \]

### Step 3: Calculate the probability of event E not occurring
The probability of event E not occurring, denoted as \( Q \), is:
\[ Q = 1 - P(E) = 1 - \frac{1}{25} = \frac{24}{25} \]

### Step 4: Use the binomial distribution
We are selecting 4 numbers, and we want to find the probability that event E occurs at least 3 times. This can be modeled using the binomial distribution.

Let \( n = 4 \) (the number of trials), \( p = \frac{1}{25} \) (the probability of success), and \( q = \frac{24}{25} \) (the probability of failure).

We need to find \( P(X \geq 3) \), which is the sum of the probabilities of event E occurring exactly 3 times and exactly 4 times:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) \]

### Step 5: Calculate \( P(X = 3) \) and \( P(X = 4) \)
Using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

1. **For \( P(X = 3) \)**:
\[ P(X = 3) = \binom{4}{3} \left(\frac{1}{25}\right)^3 \left(\frac{24}{25}\right)^1 = 4 \cdot \frac{1}{15625} \cdot \frac{24}{25} = \frac{96}{390625} \]

2. **For \( P(X = 4) \)**:
\[ P(X = 4) = \binom{4}{4} \left(\frac{1}{25}\right)^4 \left(\frac{24}{25}\right)^0 = 1 \cdot \frac{1}{390625} \cdot 1 = \frac{1}{390625} \]

### Step 6: Combine the probabilities
Now, we combine the probabilities:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) = \frac{96}{390625} + \frac{1}{390625} = \frac{97}{390625} \]

### Final Answer
The probability that event E occurs at least 3 times when 4 numbers are selected is:
\[ \frac{97}{390625} \]