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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?

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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:

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

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