# 3 balls are drawn from a bag contains 6 white balls and 4 red balls, what is the probability that 2 balls are white and 1 ball is red?

A bag contains 6 white balls and 4 red balls. If 3 balls are drawn one by one with replacement, then what is the probability that 2 balls are white and 1 ball is red?

$$\frac{18}{125}$$

What I did Probability of getting a white ball= $$6/10=3/5$$ Probability of getting a red ball= $$4/10=2/5$$ Probability of getting 2 balls white and 1 ball red = $$6/10*6/10*4/10=18/125$$

But the answer is $$\frac{54}{125}$$. Why are we multiplying it by $$3$$? Please someone elaborate this part

This is a gmat exam question.

• Look at it this way, if you don't multiply by three, then your answer is the probability that we pick $2$ white balls and $1$ red ball in that order. – WaveX Sep 9 '17 at 14:35

This is a binomial experiment with $P(W)=\frac{6}{10}=\frac{3}{5}$. Apply the formula : $$f(2)=C_2^3\cdot \left(\frac{3}{5}\right)^2\cdot \frac{2}{5}=\frac{54}{125}.$$