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Independent and Dependent Events

Example 10: A box contains 7 balls of which 4 are blue and 3 are red.
a. Suppose two balls are drawn successfully from this box with replacement, how many blue balls can you expect to draw in the first draw, and in the second draw?
b. Suppose two balls are drawn successfully from this box without replacement, how many blue balls can you expect to draw in the first draw, and in the second draw?

1 answer

To solve this problem, let's analyze each part step by step.

Part (a): Drawing with Replacement

When drawing with replacement, the composition of the box remains the same for each draw since the drawn ball is put back into the box.

  1. Probability of drawing a blue ball:

    • There are 4 blue balls and 3 red balls, which gives a total of 7 balls.
    • The probability \( P(\text{Blue}) = \frac{4}{7} \).

  2. Expected number of blue balls drawn:

    • Each draw is an independent event with the same probability.
    • The expected value of drawing a blue ball in one draw is simply the probability of drawing that blue ball.
    • Hence, the expected number for the first and the second draws is the same.
    • Therefore, for both draws:

    \[ \text{Expected Blue Balls (1st Draw)} = P(\text{Blue}) = \frac{4}{7} \] \[ \text{Expected Blue Balls (2nd Draw)} = P(\text{Blue}) = \frac{4}{7} \]

Part (b): Drawing without Replacement

When drawing without replacement, the composition of the box changes after each draw.

  1. For the first draw:

    • The probability of drawing a blue ball remains the same as in the previous case: \[ P(\text{Blue on 1st Draw}) = \frac{4}{7} \]

  2. For the second draw:

    • If the first ball drawn was blue (which happens with probability \( \frac{4}{7} \)), there would be 3 blue balls and 3 red balls left in the box.
    • If the first ball drawn was red (which happens with probability \( \frac{3}{7} \)), there would still be 4 blue balls but now only 2 red balls left in the box.

    So, we need to calculate the expected value for the second draw based on the first draw:

    • If the first was blue: \[ P(\text{Blue on 2nd Draw | 1st was Blue}) = \frac{3}{6} = \frac{1}{2} \]
    • If the first was red: \[ P(\text{Blue on 2nd Draw | 1st was Red}) = \frac{4}{6} = \frac{2}{3} \]

  3. Overall Probability of Blue on 2nd Draw:

    • Using the law of total probability: \[ P(\text{Blue on 2nd Draw}) = P(\text{Blue on 2nd Draw | 1st was Blue}) \cdot P(\text{1st was Blue}) + P(\text{Blue on 2nd Draw | 1st was Red}) \cdot P(\text{1st was Red}) \] \[ = \left( \frac{1}{2} \cdot \frac{4}{7} \right) + \left( \frac{2}{3} \cdot \frac{3}{7} \right) \]

    Simplifying each term:

    • First term: \( \frac{1}{2} \cdot \frac{4}{7} = \frac{2}{7} \)
    • Second term: \( \frac{2}{3} \cdot \frac{3}{7} = \frac{2}{7} \)

    So the overall probability: \[ P(\text{Blue on 2nd Draw}) = \frac{2}{7} + \frac{2}{7} = \frac{4}{7} \]

Therefore, the expected number of blue balls drawn is:

  • Expected Blue Balls (1st Draw): \( \frac{4}{7} \)
  • Expected Blue Balls (2nd Draw): \( \frac{4}{7} \)

Summary

  • With Replacement:

    • Expected Blue Balls (1st Draw) = \( \frac{4}{7} \)
    • Expected Blue Balls (2nd Draw) = \( \frac{4}{7} \)

  • Without Replacement:

    • Expected Blue Balls (1st Draw) = \( \frac{4}{7} \)
    • Expected Blue Balls (2nd Draw) = \( \frac{4}{7} \)