Question

A bag contains \( 5 \) red balls, \( 4 \) black balls, and \( 3 \) white balls. Then the number of ways of selecting three balls at random that contains at least one white ball is

• A. \( 220 \)
• B. \( 210 \)
• C. \( 180 \)
• D. \( 136 \)
• E. \( 74 \)

Hint:

For phrases like “at least one”, the complement method is usually the fastest approach: count total cases first, then subtract the cases with none.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a combination problem with a constraint. The phrase "at least one" suggests that it might be easier to use the principle of complementary counting.
Step 2: Key Formula or Approach:
The number of ways to select \(r\) items from a set of \(n\) items is given by the combination formula \(C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\).
The complementary counting approach is:
(Number of ways to have at least one white ball) = (Total number of ways to select 3 balls) - (Number of ways to select 3 balls with NO white balls).
Step 3: Detailed Explanation:
Total number of balls in the bag = 5 (Red) + 4 (Black) + 3 (White) = 12 balls.
1. Calculate the total number of ways to select 3 balls from 12:
This is the total number of possible combinations without any restrictions.
\[ \text{Total ways} = C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220 \] 2. Calculate the number of ways to select 3 balls with no white balls:
This means we select 3 balls only from the non-white balls.
Number of non-white balls = 5 (Red) + 4 (Black) = 9 balls.
\[ \text{Ways with no white balls} = C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \] 3. Calculate the number of ways with at least one white ball:
\[ \text{Required ways} = \text{Total ways} - \text{Ways with no white balls} \] \[ \text{Required ways} = 220 - 84 = 136 \] Step 4: Final Answer:
The number of ways of selecting three balls containing at least one white ball is 136. Therefore, option (D) is correct.