Drawing black and white balls

Question

There are two boxes: first with $5$ white and $3$ black balls, and second with $2$ white and $2$ black balls. We draw one ball from the boxes $I$ and $II$, write down colors, return them to the boxes and repeat the whole process 10 times. Compute probability that at least $2$ times drawn balls were of the same color.

So:

$A -$ at least two times drawn balls were of the same color

$A' -$ drawn balls were of the same color zero or one time $(A' = A'_0 \cup A'_1)$

$\Bbb P(A'_0) = \sum\limits_{i=0}^{10} {10 \choose i}(\frac{5}{8})^i(\frac{1}{2})^i(\frac{3}{8})^{10-i}(\frac{1}{2})^{10-i}$

And I guess this part is correct. But:

$\Bbb P(A'_1)={10 \choose 1}[\frac{5}{8}\cdot\frac{1}{2}+\frac{3}{8}\cdot\frac{1}{2}]\sum\limits_{i=0}^{9} {9 \choose i}(\frac{5}{8})^i(\frac{1}{2})^i(\frac{3}{8})^{9-i}(\frac{1}{2})^{9-i}$

I'm not really convinced that it is correct. I started with choosing one place out of $10$ for balls of the same color (white/white or black/black), and once I have it I want to have both balls in each pair to be of different color, so it is basically $A'_0$ but for drawing $9$ times instead of $10$. Am I correct?

Answer 1

Assuming your symbol $\ A_i'\ $ represents the event that the pair of balls drawn were the same colour exactly $\ i\ $ times out of $\ 10\ $ (rather than the complement of that event, which I would have found much less confusing), both your expressions for $\ \mathbb{P}\big(A_0'\big)\ $ and $\ \mathbb{P}\big(A_1'\big)\ $ are correct, but can be obtained more simply by following the procedure explained in Sam OT's answer.

You have \begin{align} \mathbb{P}\big(A_0'\big)&=\sum_{i=0}^{10} {10 \choose i}\Big(\frac{5}{8}\Big)^i\Big(\frac{1}{2}\Big)^i\Big(\frac{3}{8}\Big)^{10-i}\Big(\frac{1}{2}\Big)^{10-i}\\ &=\Big(\frac{1}{2}\Big)^{10}\sum_{i=0}^{10} {10 \choose i}\Big(\frac{5}{8}\Big)^i\Big(\frac{3}{8}\Big)^{10-i}\\ &=\Big(\frac{1}{2}\Big)^{10}\ , \end{align} which is correct, and \begin{align} \mathbb{P}\big(A_1'\big)&={10\choose1}\Bigg[\frac{5}{8} \cdot\frac{1}{2}+\frac{5}{8} \cdot\frac{1}{2}\Bigg]\\ &\hspace{2em}\times\sum_{i=0}^9 {9 \choose i}\Big(\frac{5}{8}\Big)^i\Big(\frac{1}{2}\Big)^i\Big(\frac{3}{8}\Big)^{9-i}\Big(\frac{1}{2}\Big)^{9-i}\\ &={10\choose1}\cdot\frac{1}{2}\cdot\Big(\frac{1}{2}\Big)^9\sum_{i=0}^9 {9 \choose i}\Big(\frac{5}{8}\Big)^i\Big(\frac{3}{8}\Big)^{9-i}\\ &={10\choose1}\Big(\frac{1}{2}\Big)^{10}\ , \end{align} which is also correct.

Answer 2

If I understand the situation, first, the probability of coincidence in color for any single draw is, WW or BB, then (5/8)(2/4) + (3/8)(2/4) = 1/2. Now, the number of coincidences in 10 trials is binomial of n=10 and p=0.5, then

P(at least twice) = $\sum_{i=2}^{10} \binom{10}{i} 0.5^i 0.5^{10-i} = 0.5^{10}\sum_{i=2}^{10} \binom{10}{i}$.

Your approach is somewhat similar, but I can't explain exactly your steps.

Answer 3

You don't need all these big sums. It is actually, perhaps surprisingly, very simple. Draw a ball from the first box. Regardless of which colour it is, precisely half the balls in the second box are of the same colour. So, the probability of getting the same colour is precisely $0.5 = 50%$, every time. What's pretty cool is that this is true no matter the distribution of the balls in teh first box!

Now the number is just $S \sim \operatorname{Bin}(n, 1/2)$, where $n = 10$ is the number of draws. You seem to know about these sorts of distributions, so hopefully you can take it from here!

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Ok, the formula has been given in a different answer, so I may as well give it here for completeness. In general, $$ \Pr(S = k) = \binom nk \cdot 2^{-k}. $$ Now just sum over the relevant parts: $$ \textstyle \Pr(S \ge 2) = \sum_{k=2}^{10} \binom{10}{k} 2^{-10} = 1 - \sum_{k=0}^1 \binom{10}{k} 2^{-10} = 1 - \Pr(S \le 1).$$