Question

A ball is drawn from an urn containing three white and three black balls. After the ball is drawn, it is then replaced and another ball is drawn. This goes on indefinitely. What is the probability that of the first four balls drawn, exactly two are white?

Short answer

The probability of drawing exactly 2 white balls in the first four draws is \(\frac{3}{8}\).

Step-by-step solution

Define the probabilities for drawing one white or black ball

Let's define the probabilities of drawing one white or one black ball. There's a total of 6 balls in the urn, with 3 white balls and 3 black balls. The probability of drawing one white ball is the number of white balls divided by the total number of balls: \(P(W) = \frac{3}{6} = \frac{1}{2}\) Similarly, the probability of drawing one black ball is the number of black balls divided by the total number of balls: \(P(B) = \frac{3}{6} = \frac{1}{2}\)

Using the combination formula

Now, let's use the combination formula to find the possible ways we can draw exactly 2 white balls and 2 black balls in 4 draws. The combination formula is given by \(C(n, r) = \frac{n!}{r!(n-r)!}\), where n is the total number of items and r is the number of items to be chosen. Here, we choose 2 white balls from 4 draws, so n=4 and r=2. We calculate \(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4\cdot3\cdot2\cdot1}{2\cdot1\cdot2\cdot1} = 6\)

Calculate the probability

To calculate the probability of exactly two white balls and two black balls in four draws, we need to multiply the probabilities of drawing white and black balls with the number of possible ways. So, \(P(\text{exactly 2 white balls}) = C(4, 2) \times P(W)^2 \times P(B)^2\) Substituting the values, we get: \(P(\text{exactly 2 white}) = 6 \times (\frac{1}{2})^2 \times (\frac{1}{2})^2 = 6 \times \frac{1}{16} = \frac{3}{8}\) Thus, the probability of drawing exactly 2 white balls in the first four draws is \(\frac{3}{8}\).