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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?

My answer came out as

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

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.

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    $\begingroup$ 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. $\endgroup$ – WaveX Sep 9 '17 at 14:35
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You have to multiply by three because you can draw the red as either the first, the second or the third ball.

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

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