Question:
A bag has 5 blue balls and 15 red balls. Three balls are drawn at random from the bag simultaneously. Then the probability that none of the chosen balls is blue equals
A bag has 5 blue balls and 15 red balls. Three balls are drawn at random from the bag simultaneously. Then the probability that none of the chosen balls is blue equals
Updated On: Apr 15, 2025
- \(\frac{75}{152}\)
- \(\frac{91}{228}\)
- \(\frac{27}{64}\)
- \(\frac{273}{800}\)
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The Correct Option is B
Solution and Explanation
1. Total Number of Balls: - Total balls = \( 5 \, (\text{blue}) + 15 \, (\text{red}) = 20 \).
2. Favorable Outcomes: - If none of the balls is blue, all three chosen balls must be red. - The number of ways to choose 3 red balls from 15 red balls: \[ \binom{15}{3} = \frac{15 \cdot 14 \cdot 13}{3 \cdot 2 \cdot 1} = 455. \] 3. Total Outcomes: - The total number of ways to choose 3 balls from 20: \[ \binom{20}{3} = \frac{20 \cdot 19 \cdot 18}{3 \cdot 2 \cdot 1} = 1140. \] 4. Probability: - The probability is: \[ P = \frac{\binom{15}{3}}{\binom{20}{3}} = \frac{455}{1140} = \frac{91}{228}. \]
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