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I'm having trouble with the following problem

There are 4 different colored balls: Red, Green, Black, Blue what is the probability that I will get 2 of the same colored balls by drawing 4 balls with replacement.

I attempted to solve this problem by splitting the question into two parts. First I calculated the probability of drawing 4 balls where I have 2 of one color and 2 of another color. Call this event $A$. $$P(A) = \frac{4 \times 1 \times 3 \times 1 \times \binom{4}{2}}{ 4^4}= \frac{72}{4^4}$$

Next I calculated the probability where 2 balls are the same color and the other 2 are different. Call this event $B$ $$P(B)= \frac{4\times1\times3\times2\times4P2}{4^4}$$ I understand that if I sum these two probabilities I should get the answer but for some reason the event B was not calculated correctly. Any feedback and help is appreciated.

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  • $\begingroup$ You draw only 2 balls. $\endgroup$ – true blue anil Oct 22 '15 at 3:35
  • $\begingroup$ Edited the question as I missed a key detail, yes I am drawing 4 balls. $\endgroup$ – Daniel Oct 22 '15 at 3:59
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Hint: First try to think about how many color combinations you can have. Next try to think how you can arrange four balls. When you arrange the balls you have to think carefully about same colored balls.

Based on your attempt, I assume that we draw four balls even though it is not specified in your question.

Following your logic,

$$P(A) = \frac{\binom{4}{2}\frac{4!}{2!2!}}{4^4}$$

$$P(B) = \frac{\binom{4}{1}\binom{3}{2}\frac{4!}{2!}}{4^4}$$

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  • $\begingroup$ Did I approach the question correctly though? I'm having trouble distinguishing whether drawing RRBB is the same as drawing BBRR in this question. $\endgroup$ – Daniel Oct 22 '15 at 3:59
  • $\begingroup$ Based on your calculation of denominator as $4^4$, we treat RRBB and BBRR differently. $\endgroup$ – Statbot Oct 22 '15 at 4:04
  • $\begingroup$ Do you think you could elaborate on how you found $P(B)$ I understand that $\frac{4!}{2!}$ is the number of ways you can arrange the balls but how did you come up with $\binom{4}{3}$ $\endgroup$ – Daniel Oct 22 '15 at 4:09
  • $\begingroup$ Why would it be necessary to choose 3 colors from the four choices if you need two to be the same and the other two to be different sorry I'm having a hard time wrapping my head around this. $\endgroup$ – Daniel Oct 22 '15 at 4:31
  • $\begingroup$ Sorry, I think I made mistake before. First we choose one color from four that will be used twice and then you choose two more colors from three remaining choices. $\endgroup$ – Statbot Oct 22 '15 at 4:36
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Where does it say that you have to draw 4 balls ???

I take it that there are a very large number of balls of each color, and we draw only 2 balls.

Thus $Pr = 4\times \dfrac14\times \dfrac14 = \dfrac14$

PS:

Even with your clarification that 4 balls are drawn, the question is ambiguous. Is it exactly 2 of the same color or at least 2 of the same color, and is 2 of one color and 2 of another color to be counted ?

Anyway, I am giving the "partial" answers which you can sew together.

2-1-1 of a kind: $\dbinom41\dbinom32 \dfrac{4!}{2!}$ ways

2-2 of a kind: $\dbinom42 \dfrac{4!}{2!2!}$ ways

3-1 of a kind: $\dbinom41 \dbinom31 \dfrac{4!}{3!}$ ways

4 of a kind: $\dbinom41$ ways

Divide by $4^4$ to get probabilities

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  • $\begingroup$ Sorry for the confusion I forgot to add that piece of information yes I'm drawing 4 balls. $\endgroup$ – Daniel Oct 22 '15 at 4:00

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