A box contains tickets numbered from 1 to 20. If 3 tickets are drawn one by one with replacement then the probability of getting prime number exactly 2 times is

Options

A. `(36)/(125)`
B. `(12)/(125)`
C. `(1)/(125)`
D. `(4)/(125)`

Answer

To solve the problem of finding the probability of drawing a prime number exactly 2 times when drawing 3 tickets from a box containing tickets numbered from 1 to 20 (with replacement), we can follow these steps:

### Step 1: Identify the Prime Numbers
The prime numbers between 1 and 20 are:
- 2, 3, 5, 7, 11, 13, 17, 19

Thus, there are a total of 8 prime numbers.

### Step 2: Calculate the Total Number of Tickets
The total number of tickets is 20 (from 1 to 20).

### Step 3: Calculate the Probability of Drawing a Prime Number
The probability \( P(\text{Prime}) \) of drawing a prime number is given by:
\[ P(\text{Prime}) = \frac{\text{Number of Prime Numbers}}{\text{Total Number of Tickets}} = \frac{8}{20} = \frac{2}{5} \]

### Step 4: Calculate the Probability of Not Drawing a Prime Number
The probability \( P(\text{Not Prime}) \) of not drawing a prime number is:
\[ P(\text{Not Prime}) = 1 - P(\text{Prime}) = 1 - \frac{2}{5} = \frac{3}{5} \]

### Step 5: Use the Binomial Probability Formula
We want to find the probability of getting exactly 2 prime numbers in 3 draws. This can be modeled using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n = 3 \) (the number of trials),
- \( k = 2 \) (the number of successes),
- \( p = \frac{2}{5} \) (the probability of success),
- \( 1-p = \frac{3}{5} \) (the probability of failure).

### Step 6: Calculate the Binomial Coefficient
The binomial coefficient \( \binom{3}{2} \) is calculated as follows:
\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \]

### Step 7: Substitute Values into the Binomial Formula
Substituting the values into the formula:
\[ P(X = 2) = \binom{3}{2} \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)^{3-2} \]
\[ P(X = 2) = 3 \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)^1 \]

### Step 8: Calculate the Probability
Now we calculate:
\[ P(X = 2) = 3 \left(\frac{4}{25}\right) \left(\frac{3}{5}\right) = 3 \times \frac{4 \times 3}{25 \times 5} = 3 \times \frac{12}{125} = \frac{36}{125} \]

### Conclusion
Thus, the probability of getting a prime number exactly 2 times when drawing 3 tickets with replacement is:
\[ \frac{36}{125} \]