A bag contains 4 red and 3 blue balls. If two balls are drawn from this bag at random, what is the probability that they are of different colour?

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DSSSB Pharmacist 2014: Previous Year Paper
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  1. \(\frac{2}{7}\)
  2. \(\frac{4}{7}\)
  3. \(\frac{3}{7}\)
  4. \(\frac{5}{7}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{4}{7}\)

Detailed Solution

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Given:

A bag contains 4 red and 3 blue balls. If two balls are drawn from this bag at random.

Formula used:

The probability of an event = Number of favorable outcomes / Total number of outcomes

Calculation:

  1. Total number of ways to draw 2 balls out of 7:

    \(7C_2 = \frac{7!}{2!(7-2)!} = \frac{7 × 6}{2 × 1} = 21\)

  2. Number of ways to draw 1 red and 1 blue ball:
    • Number of ways to draw 1 red ball out of 4 = \(4C_1 = 4\)
    • Number of ways to draw 1 blue ball out of 3 = \(3C_1 = 3\)

    Therefore, the number of ways to draw 1 red and 1 blue ball = 4 × 3 = 12

  3. Hence, the probability that the two balls drawn are of different colors:

    Probability = Number of favorable outcomes / Total number of outcomes = 12/21 = 4/7

The probability that the two balls drawn are of different colors is 4/7

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