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A drawer holds $4$ red hats and $4$ blue hats. What is the probability of getting exactly $3$ red hats or $3$ blue hats when taking out $4$ hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?

When I come across the question (I saw this questions two times in a week), I tried to solve in this way-

probability of getting exactly $3$ red hats is

$$\frac{{4\choose 3}{4\choose 1}}{{8\choose 4}} = \frac{16}{70}$$

probability of getting exactly $3$ blue hats is

$$\frac{{4\choose 3}{4\choose 1}}{{8\choose 4}} = \frac{16}{70}$$

So, P(exactly $3$ red or exactly $3$ blue ) = $\frac{32}{70}$

But answer of this problem is $\frac{1}{2}$. What is my mistake?

Mahmudul Hasan
  • 711

1 Answers1

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Since the hats are being replaced after each draw, the probability of pulling a red hat is $4/8$ on every draw, and the probability of pulling a blue hat is also $4/8$. There are ${4\choose 3}$ ways to pull 3 red hats in 4 draws, so the probability of choosing exactly 3 red hats is \begin{equation} {4\choose 3}\left(\frac{4}{8}\right)^3\left(\frac{4}{8}\right)^1 = \frac{4!}{3!1!}\frac{1}{16} = \frac{1}{4}. \end{equation} Similarly, the probability of pulling exactly 3 blue hats is $1/4$, so the probability of pulling exactly 3 red hats or exactly 3 blue hats is the sum, $1/2$.

Your strategy went awry because the formulas you're using are for drawing without replacement.

211792
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