Question

From 4 red balls, 2 white balls and 4 black balls, four balls are selected. The probability of getting 2 red balls is:

• A. \( \frac{7}{21} \)
• B. \( \frac{8}{21} \)
• C. \( \frac{9}{21} \)
• D. \( \frac{10}{21} \)
• E. \( \frac{11}{21} \)

Hint:

In multi-color ball problems, group the balls into "Target Color" and "Others" to simplify the combination calculation. It prevents confusion between multiple non-target categories.

Answer 1

The Correct Option is C

Concept: When selecting multiple items, we use combinations \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). To find the probability of getting exactly 2 red balls when 4 are drawn, we calculate the ways to choose 2 red balls from the red group and 2 non-red balls from the remaining group, then divide by the total ways to draw 4 balls.

Step 1: Calculating total ways and favorable ways.
Total balls = \( 4 (\text{red}) + 2 (\text{white}) + 4 (\text{black}) = 10 \). Total ways to select 4 balls = \( \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \). Favorable ways (2 red and 2 non-red): - Ways to pick 2 red from 4 = \( \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \). - Ways to pick 2 non-red from 6 (white + black) = \( \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \). Total favorable ways = \( 6 \times 15 = 90 \).

Step 2: Calculating the probability.
\[ P = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{90}{210} \] Simplifying the fraction: \[ P = \frac{9}{21} \]