A bag has 10 balls. Six ball are drawn in an attempt and replaced. Then another draw of 5 balls is made from the bag. The probability that exactly two balls are common to both the draw is a.`5//21` b. `2//21` c. `7//21` d. `3//21`

Options

A. `5//21`
B. `2//21`
C. `7//21`
D. `3//21`

Explanation

To solve the problem of finding the probability that exactly two balls are common to both draws from a bag containing 10 balls, we can follow these steps:

### Step 1: Understand the Problem
We have a bag with 10 balls. In the first draw, we draw 6 balls and replace them. In the second draw, we draw 5 balls. We need to find the probability that exactly 2 balls are common to both draws.

### Step 2: Total Outcomes for the Second Draw
The total number of ways to choose 5 balls from 10 is given by the combination formula:
\[ \text{Total outcomes} = \binom{10}{5} \]

### Step 3: Favorable Outcomes
To find the favorable outcomes where exactly 2 balls are common in both draws, we can break it down into two parts:
1. Choose 2 balls from the 6 balls drawn in the first attempt.
2. Choose the remaining 3 balls from the 4 balls that were not drawn in the first attempt.

The number of ways to choose 2 balls from the 6 drawn is:
\[ \text{Ways to choose 2 from 6} = \binom{6}{2} \]

The number of ways to choose 3 balls from the remaining 4 (10 total - 6 drawn = 4 remaining) is:
\[ \text{Ways to choose 3 from 4} = \binom{4}{3} \]

### Step 4: Calculate Favorable Outcomes
Now we can calculate the total number of favorable outcomes:
\[ \text{Favorable outcomes} = \binom{6}{2} \times \binom{4}{3} \]

### Step 5: Calculate the Probability
The probability of exactly 2 balls being common is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{exactly 2 common}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{\binom{6}{2} \times \binom{4}{3}}{\binom{10}{5}} \]

### Step 6: Substitute Values and Simplify
Now we substitute the values of the combinations:
- \(\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15\)
- \(\binom{4}{3} = \frac{4!}{3!(4-3)!} = 4\)
- \(\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 252\)

Now we can calculate:
\[ \text{Favorable outcomes} = 15 \times 4 = 60 \]
\[ P(\text{exactly 2 common}) = \frac{60}{252} \]

### Step 7: Simplify the Probability
Now we simplify \(\frac{60}{252}\):
\[ \frac{60}{252} = \frac{5}{21} \]

### Final Answer
Thus, the probability that exactly two balls are common to both draws is:
\[ \frac{5}{21} \]