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

Options

A. `2.4832xx10^(-4)`
B. `2.481xx10^(-4)`
C. `2.47xx10^(-4)`
D. none of these

Answer

To solve the problem step by step, we need to find the probability that the event \( E \) occurs at least 3 times when selecting 4 two-digit numbers (from 00 to 99) with replacement. The event \( E \) occurs if the product of the two digits of the selected number equals 18.

### Step 1: Identify the two-digit numbers where the product of the digits equals 18.

The two-digit numbers can be represented as \( ab \), where \( a \) is the tens digit and \( b \) is the units digit. We need to find pairs \( (a, b) \) such that \( a \times b = 18 \).

The pairs that satisfy this condition are:
- \( (2, 9) \) → 29
- \( (3, 6) \) → 36
- \( (6, 3) \) → 63
- \( (9, 2) \) → 92

Thus, the valid two-digit numbers are 29, 36, 63, and 92. This gives us a total of 4 favorable outcomes.

### Step 2: Calculate the probability \( p \) of event \( E \).

There are a total of 100 two-digit numbers (from 00 to 99). Therefore, the probability \( p \) that a randomly selected number satisfies event \( E \) is:

\[ p = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{100} = \frac{1}{25} \]

### Step 3: Calculate the probability \( q \) of the complement event (not \( E \)).

The probability \( q \) that the event \( E \) does not occur is:

\[ q = 1 - p = 1 - \frac{1}{25} = \frac{24}{25} \]

### Step 4: Use the binomial distribution to find the probability of event \( E \) occurring at least 3 times.

We will use the binomial distribution formula:

\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

where \( n = 4 \) (the number of trials), \( r \) is the number of successes, \( p \) is the probability of success, and \( q \) is the probability of failure.

We need to calculate \( P(X \geq 3) \), which is:

\[ P(X \geq 3) = P(X = 3) + P(X = 4) \]

### Step 5: Calculate \( P(X = 3) \).

Using the binomial formula:

\[ P(X = 3) = \binom{4}{3} \left(\frac{1}{25}\right)^3 \left(\frac{24}{25}\right)^{4-3} \]

Calculating \( \binom{4}{3} = 4 \):

\[ P(X = 3) = 4 \cdot \left(\frac{1}{25}\right)^3 \cdot \left(\frac{24}{25}\right)^1 \]

\[ = 4 \cdot \frac{1}{15625} \cdot \frac{24}{25} = \frac{96}{390625} \]

### Step 6: Calculate \( P(X = 4) \).

Using the binomial formula:

\[ P(X = 4) = \binom{4}{4} \left(\frac{1}{25}\right)^4 \left(\frac{24}{25}\right)^{0} \]

Calculating \( \binom{4}{4} = 1 \):

\[ P(X = 4) = 1 \cdot \left(\frac{1}{25}\right)^4 = \frac{1}{390625} \]

### Step 7: Combine the probabilities.

Now we combine the probabilities:

\[ P(X \geq 3) = P(X = 3) + P(X = 4) = \frac{96}{390625} + \frac{1}{390625} = \frac{97}{390625} \]

### Step 8: Final result.

The final probability that event \( E \) occurs at least 3 times when selecting 4 numbers is:

\[ P(X \geq 3) = \frac{97}{390625} \approx 0.00024832 \]