(a) There are 3 red and 5 black balls in one box and 6 red and 4 white balls in another. If you pick a box at random, and then pick a ball from it at random, what is the probability that it is red? Black? White? That it is either red or white?

(b) Suppose the first ball selected is red and is not replaced before a second ball

is drawn. What is the probability that the second ball is red also?

(c) If both balls are red, what is the probability that they both came from the same box?

There are 3 red and 5 black balls in one box and 6 red and 4 white balls in another box.

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

There are total 8balls in first box out of which 3 are red and total 10red balls are in the second out of which 6 are red and each box has a probability $\frac{1}{2}$of getting picked.

Find the probability of picking a red ball from any box.

$$\begin{gathered} P\left(Red\right)=\frac{1}{2}\times \frac{3}{8}I\frac{1}{2}\times \frac{6}{10} \\ =\frac{39}{80} \end{gathered}$$

Find the probability of picking a black ball from any box.

$$\begin{gathered} P\left(Black\right)=\frac{1}{2}+\frac{5}{8}\times \frac{0}{10} \\ =\frac{5}{16} \end{gathered}$$

Find the probability of picking a white ball from any box.

$$\begin{gathered} P\left(White\right)=\frac{1}{2}\times \frac{0}{8}I\frac{1}{2}\times \frac{4}{10} \\ =\frac{1}{5} \end{gathered}$$

Find the probability of pickingeither red or white ball from any box.

$$\begin{gathered} P\left(Red+White\right)=\frac{1}{2}\times \frac{3}{8}+\frac{1}{2}\times \left(\frac{6}{10}+\frac{4}{10}\right) \\ =\frac{3}{16}+\frac{1}{2} \\ =\frac{11}{16} \end{gathered}$$

Find The probability that the second ball is also red when first ball is redusing the Bayes theorem, $P_A\left(B\right)=\frac{P\left(AB\right)}{P\left(A\right)}$.

$$\begin{gathered} P\left(BothRedIOneRed\right)=\frac{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{8}}\times {\displaystyle \frac{1}{2}}+\times {\displaystyle \frac{6}{10}}\times {\displaystyle \frac{5}{9}}}{{\displaystyle \frac{39}{80}}} \\ =\frac{370}{819} \end{gathered}$$

Thus, the desired probability is $\frac{370}{819}$.

Find the probability of getting both reds.

$\mathit{P}\left(BothReds\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{8}}\mathbf{\times }\frac{\mathbf{2}}{\mathbf{7}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{8}}\mathbf{\times }\frac{\mathbf{6}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{6}}{\mathbf{10}}\mathbf{\times }\frac{\mathbf{3}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{6}}{\mathbf{10}}\mathbf{\times }\frac{\mathbf{5}}{\mathbf{9}}$

Find the probability of getting both reds from same box.

$\mathit{P}\left(BothRedsfromsamebox\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{8}}\mathbf{\times }\frac{\mathbf{2}}{\mathbf{7}}\mathbf{+}\frac{\mathbf{6}}{\mathbf{10}}\mathbf{\times }\frac{\mathbf{5}}{\mathbf{9}}$

Find the probability that they both came from the same box, when both balls are red

Using the Bayes theorem, $P_A\left(B\right)=\frac{P\left(AB\right)}{P\left(A\right)}$.

role="math" localid="1654839283298" $$\begin{gathered} \mathit{P}\left(BothRedsfromsameboxIBothRed\right)\mathbf{=}\frac{{\displaystyle \frac{3}{8}\times \frac{2}{7}+\frac{6}{10}\times \frac{5}{9}}}{{\displaystyle \frac{3}{8}\times \frac{2}{7}+\frac{3}{8}\times \frac{6}{10}+\frac{6}{10}\times \frac{3}{8}+\frac{6}{10}\times \frac{5}{9}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\approx }\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$