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?
Best Answer
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.