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: Find the total number of balls.
The bag contains: \[ 5 \text{ red} + 4 \text{ black} + 3 \text{ white} = 12 \text{ balls} \] We need to select \( 3 \) balls such that at least one of them is white.

Step 2: Use the complementary counting method.
Instead of counting selections with at least one white ball directly, it is easier to count: \[ \text{Total selections of 3 balls} - \text{Selections with no white ball} \] This method is simpler and avoids multiple cases.

Step 3: Count total selections of 3 balls.
From \( 12 \) balls, the number of ways to choose any \( 3 \) balls is \[ {}^{12}C_3=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} \] \[ =220 \]

Step 4: Count selections with no white ball.
If no white ball is selected, then all \( 3 \) balls must come from red and black balls only.
Number of non-white balls: \[ 5+4=9 \] So, the number of ways to choose \( 3 \) balls from these \( 9 \) balls is \[ {}^{9}C_3=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} \] \[ =84 \]

Step 5: Subtract to get the required number.
Hence, the number of selections containing at least one white ball is \[ {}^{12}C_3-{}^{9}C_3 \] \[ =220-84 \] \[ =136 \]

Step 6: Verify the interpretation.
The phrase “at least one white ball” includes: \[ 1 \text{ white},\quad 2 \text{ white},\quad \text{or } 3 \text{ white} \] All such cases are automatically included in the complement method, so the answer is complete.

Step 7: Final conclusion.
Therefore, the required number of ways is \[ \boxed{136} \] Hence, the correct option is \[ \boxed{(4)\ 136} \]