A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
- \(\frac{2}{91}\)
- \(\frac{3}{22}\)
- \(\frac{5}{71}\)
- \(\frac{7}{51}\)
The Correct Option is A
Solution and Explanation
Probability Analysis
Total number of balls = 4 (white) + 5 (red) + 6 (blue) = 15 balls.
The number of ways to choose 3 balls from 15 is given by:
\( \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \).
The number of ways to choose 3 red balls from 5 is given by:
\( \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \).
The probability of drawing 3 red balls is:
\( \frac{\binom{5}{3}}{\binom{15}{3}} = \frac{10}{455} = \frac{2}{91} \).
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Let \( A_1 \): People with good health,
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