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.
Best Answer
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,
$$P(A) = \frac{\binom{4}{2}\frac{4!}{2!2!}}{4^4}$$
$$P(B) = \frac{\binom{4}{1}\binom{3}{2}\frac{4!}{2!}}{4^4}$$