91Ó°ÊÓ

There are 6Ns in first box and 2Ns and 3Ps in second 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.

Draw the tree diagram depicting the transistor is taken from each box and corresponding probability.

Draw the tree diagram depicting the transistor is taken from each box and corresponding probability.

Find the probability that selected transistor comes from first box using the Bayes Theorem and the tree diagram.

$$\begin{gathered} P\left(Ntransostors\overline{)6NBox}\right)=\frac{P\left(6NBoxandNTransistor\right)}{P\left(NTransistor\right)} \\ =\frac{{\displaystyle \frac{1}{2}}×{\displaystyle \frac{6}{6}}}{{\displaystyle \frac{1}{2}}×{\displaystyle \frac{6}{6}}+{\displaystyle \frac{1}{2}}×{\displaystyle \frac{2}{5}}} \\ =\frac{5}{7} \end{gathered}$$

Find the probability that selected transistor comes from second box by subtracting the obtained probability from 1.

$$\begin{gathered} P\left(Ntransistors\overline{)2N,3P}Box\right)=1-P\left(Ntransistors\overline{)6N}Box\right) \\ =1-\frac{5}{7} \\ =\frac{2}{7} \end{gathered}$$

Find the probability that another transistor is picked from the same box as the first and it is also an N by multiplying the probability of picking the second N transistor from each box to the obtained probability corresponding to the boxes.

$$\begin{gathered} P\left(SecondNtransistor\right)=\frac{5}{7}×\frac{5}{5}+\frac{2}{7}×\frac{2}{4} \\ =\frac{11}{14} \end{gathered}$$