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A box contains 2 green apples and 2 red apples. Apples are taken from the box, one at a time, withoutreplacement. When both red apples have been taken, the process stops. The random variable X is thenumber of apples which have been taken when the process stops.(i) Show that P(X = 3) =

Question

A box contains 2 green apples and 2 red apples. Apples are taken from the box, one at a time, withoutreplacement. When both red apples have been taken, the process stops. The random variable X is thenumber of apples which have been taken when the process stops.(i) Show that P(X = 3) =

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Solution

The event X = 3 means that the two red apples are taken out in the first three draws. This can happen in three ways:

  1. The first apple is red, the second apple is red, and the third apple is green.
  2. The first apple is red, the second apple is green, and the third apple is red.
  3. The first apple is green, the second apple is red, and the third apple is red.

Let's calculate the probability for each case:

  1. The probability that the first apple is red is 2/4 = 1/2. After taking one red apple out, the probability that the second apple is red is 1/3. The probability that the third apple is green is then 2/2 = 1. So, the probability for this case is (1/2) * (1/3) * 1 = 1/6.

  2. The probability that the first apple is red is 2/4 = 1/2. The probability that the second apple is green is 2/3. After taking one red and one green apple out, the probability that the third apple is red is 1/2. So, the probability for this case is (1/2) * (2/3) * (1/2) = 1/6.

  3. The probability that the first apple is green is 2/4 = 1/2. After taking one green apple out, the probability that the second apple is red is 2/3. The probability that the third apple is red is then 1/2. So, the probability for this case is (1/2) * (2/3) * (1/2) = 1/6.

Therefore, the total probability that X = 3 is 1/6 + 1/6 + 1/6 = 1/2.

This problem has been solved

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