Question

A bag contains 5 white and 3 black balls, and 4 are successively drawn out and not replaced. What’s the chance of getting different colours alternatively?

• A. \(\frac{1}{6}\)
• B. \(\frac{1}{5}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{7}\)

Answer 1

The Correct Option is D

To find the probability of drawing balls with alternating colors, we must first consider the total possible ways to draw 4 balls out of 8.

Number of white balls (W)	5
Number of black balls (B)	3
Total balls	8

The sequence of drawing can be done in two patterns for alternating colors:

• White, Black, White, Black (W, B, W, B)
• Black, White, Black, White (B, W, B, W)

Total combinations for alternating colors calculation:

1. For W, B, W, B:
First ball
Choose 1st white:
\(\frac{5}{8}\)
Choose next black:
\(\frac{3}{7}\)
Choose next white:
\(\frac{4}{6}\)
Choose last black:
\(\frac{2}{5}\)

2. For B, W, B, W:
First ball
Choose 1st black:
\(\frac{3}{8}\)
Choose next white:
\(\frac{5}{7}\)
Choose next black:
\(\frac{2}{6}\)
Choose last white:
\(\frac{4}{5}\)

Total probability for pattern W, B, W, B:

\(\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} \times \frac{2}{5} = \frac{1}{14}\)

Total probability for pattern B, W, B, W:

\(\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} \times \frac{4}{5} = \frac{1}{14}\)

Total probability for alternating colors:
\(\frac{1}{14} + \frac{1}{14} = \frac{2}{14} = \frac{1}{7}\)

Hence, the probability of drawing 4 balls in alternatively different colors is \(\frac{1}{7}\).