Question:

A bag contains 4 red and 6 black balls. Two balls are drawn without replacement. The probability that both are red is:

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Without replacement → probabilities change after first draw.
Updated On: Jan 4, 2026
  • \(\frac{1}{15}\)
  • \(\frac{2}{15}\)
  • \(\frac{3}{15}\)
  • \(\frac{4}{15}\) 

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The Correct Option is C

Solution and Explanation

To find the probability that both balls drawn from the bag are red, we can follow these steps: 1. Identify Total Outcomes: - The total number of balls in the bag is 4 red + 6 black = 10 balls. - We are drawing two balls, so the total number of ways to choose 2 balls out of 10 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. \[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. Calculate Favorable Outcomes: - The number of ways to choose 2 red balls out of the 4 red balls is: \[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \] 3. Calculate the Probability: - The probability that both balls drawn are red is the ratio of the number of favorable outcomes to the total number of outcomes. \[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{6}{45} = \frac{2}{15} \] 4. Check the Correctness with Options: - The given options were \(\frac{1}{15}\), \(\frac{2}{15}\), \(\frac{3}{15}\), and \(\frac{4}{15}\). - The correct probability, based on our calculation, matches with \(\frac{2}{15}\). In conclusion, the probability that both balls drawn are red is \(\frac{2}{15}\).

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