Question

A box contains 10 balls out of which 3 are red and the rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same colour?

• A. 105
• B. 168
• C. 189
• D. 120

Answer 1

Answer: (B)

168

Answer 2

The correct option is (B): 168
To find the number of ways to draw a random sample of 6 balls from the box with the condition that at most 2 red balls are included and not all 6 balls are of the same color, we can break this problem down into cases based on the number of red balls drawn.

Given:
- Total balls = 10 (3 red + 7 blue)
- Sample size = 6 balls
- Condition: At most 2 red balls in the sample, and no sample can consist of all balls of the same color.

Cases:
1. Case 1: 0 red balls and 6 blue balls (not allowed as it contains all the same color).
2. Case 2: 1 red ball and 5 blue balls.
3. Case 3: 2 red balls and 4 blue balls.
Case 2: 1 Red Ball and 5 Blue Balls
- Choose 1 red ball from 3 red balls: $\binom{3}{1}$
- Choose 5 blue balls from 7 blue balls: $\binom{7}{5}$

Calculating:
$\binom{3}{1} = 3$
$\binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$
Total for Case 2:
$3 \times 21 = 63$

Case 3: 2 Red Balls and 4 Blue Balls
- Choose 2 red balls from 3 red balls: $\binom{3}{2}$
- Choose 4 blue balls from 7 blue balls: $\binom{7}{4}$

Calculating:
$\binom{3}{2} = 3$
$\binom{7}{4} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
Total for Case 3:
$3 \times 35 = 105$

Total Number of Ways
Adding the results from both valid cases:
$\text{Total} = 63 + 105 = 168$

Conclusion
The number of ways to draw a random sample of 6 balls, meeting the conditions specified, is 168.

So, the correct answer is 168.