# One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls

One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls. A ball is chosen at random from the first container and placed in the second. Then a ball is chosen at random from the second container and placed in first.

a. What is the probability to choose a red ball of the second container?

b. What is the probability the ball was red chosen from the first container if know that from the second selected white?

Solution for question 1:

Probability that a red ball will be chosen on first step: $\frac{6}{10}$
In this case, we will have 8 red and 3 white balls in the second container, and the probability to chose a red ball on second step is $\frac{8}{11}$.
Overall probability in this case gives $\frac{6}{10}*\frac{8}{11}$ = $\frac{48}{110}$. Let's mark it as p$_1$.

Probability that a white ball will be chosen on first step: $\frac{4}{10}$
Similarly to the previous case, we will get probability to choose a red ball from the second container is p$_2$=$\frac{4}{10}*\frac{7}{11}=\frac{28}{110}$

Because our cases do not intersect, we have total probability p=p$_1$+p$_2$=$\frac{76}{110}$=$\frac{38}{55}$.

Solutin for question 2:

Here we shall use Bayes's theorem: $$P(A|B)=\frac{P(B|A)*P(A)}{P(B)}$$

Where event A means that a red ball was chosen from the first container, B means that a white ball was chosen from the second container.

What we know:

$P(A)=\frac{6}{10}$
$P(B)=1-\frac{38}{55}=\frac{17}{55}$ (1-probability that a red ball will be chosen from the second container) $P(B|A)=\frac{6}{10}*\frac{3}{11}=\frac{9}{55}$

And here we are:

$$P(A|B)=\frac{\frac{9}{55}*\frac{6}{10}}{\frac{17}{55}}=\frac{27}{85}$$