Refer to question 41 above. If a white ball is selected, what is the probability that it came from
(i) Bag II?
(ii) Bag III?
Referring to the previous solution, using Bay's theorem, we have
$$\begin{aligned} \text{(i) }\quad P\left(E_2 / F\right) & =\frac{P\left(E_2\right) \cdot P\left(F / E_2\right)}{P\left(E_1\right) \cdot P\left(F / E_1\right)+P\left(E_2\right) \cdot P\left(F / E_2\right)+P\left(E_3\right) \cdot P\left(F / E_3\right)} \\ & =\frac{\frac{2}{6} \cdot \frac{1}{3}}{\frac{1}{6} \cdot 0+\frac{2}{6} \cdot \frac{1}{3}+\frac{3}{6} \cdot 1}=\frac{\frac{2}{18}}{\frac{2}{18}+\frac{3}{6}} \\ & =\frac{2 / 18}{\frac{2+9}{18}}=\frac{2}{11} \end{aligned}$$
$$\begin{aligned} \text{(ii)}\quad P\left(E_3 / F\right) & =\frac{P\left(E_3\right) \cdot P\left(F / E_3\right)}{P\left(E_1\right) \cdot P\left(F / E_1\right)+P\left(E_2\right) \cdot P\left(F / E_2\right)+P\left(E_3\right) \cdot P\left(F / E_3\right)} \\ & =\frac{\frac{3}{6} \cdot 1}{\frac{1}{6} \cdot 0+\frac{2}{6} \cdot \frac{1}{3}+\frac{3}{6} \cdot 1} \\ & =\frac{\frac{3}{6}}{\frac{2}{18}+\frac{3}{6}}=\frac{3 / 6}{\frac{2}{18}+\frac{9}{18}}=\frac{9}{11} \end{aligned}$$
A shopkeeper sells three types of flower seeds $A_1, A_2$ and $A_3$. They are sold as a mixture, where the proportions are $4: 4: 2$, respectively. The germination rates of the three types of seeds are $45 \%, 60 \%$ and $35 \%$. Calculate the probability̌/p>
(i) of a randomly chosen seed to germinate.
(ii) that it will not germinate given that the seed is of type $A_3$.
(iii) that it is of the type $A_2$ given that a randomly chosen seed does not germinate.
$$\begin{aligned} &\text { We have, } A_1: A_2: A_3=4: 4: 2\\ &P\left(A_1\right)=\frac{4}{10}, P\left(A_2\right)=\frac{4}{10} \text { and } P\left(A_3\right)=\frac{2}{10} \end{aligned}$$
where $A_1, A_2$ and $A_3$ denote the three types of flower seeds.
Let $E$ be the event that a seed germinates and $\bar{E}$ be the event that a seed does not germinate.
$$\therefore \quad P\left(E / A_1\right)=\frac{45}{100}, P\left(E / A_2\right)=\frac{60}{100} \text { and } P\left(E / A_3\right)=\frac{35}{100}$$
$$\begin{aligned} &\begin{aligned} \text { (i) } \therefore\quad P(E) & =P\left(A_1\right) \cdot P\left(E / A_1\right)+P\left(A_2\right) \cdot P\left(E / A_2\right)+P\left(A_3\right) \cdot P\left(E / A_3\right) \\ & =\frac{4}{10} \cdot \frac{45}{100}+\frac{4}{10} \cdot \frac{60}{100}+\frac{2}{10} \cdot \frac{35}{100} \\ & =\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49 \end{aligned} \end{aligned}$$
(ii) $P\left(\bar{E} / A_3\right)=1-P\left(E / A_3\right)=1-\frac{35}{100}=\frac{65}{100}$ [as given above]
$$\begin{aligned} &\text { (iii) }\\ &\begin{aligned} P\left(A_2 / \bar{E}\right) & =\frac{P\left(A_2\right) \cdot P\left(\bar{E} / A_2\right)}{P\left(A_1\right) \cdot P\left(\bar{E} / A_1\right)+P\left(A_2\right) \cdot P\left(\bar{E} / A_2\right)+P\left(A_3\right) \cdot P\left(\bar{E} / A_3\right)} \\ & =\frac{\frac{4}{10} \cdot \frac{40}{100}}{\frac{4}{10} \cdot \frac{55}{100}+\frac{4}{10} \cdot \frac{40}{100}+\frac{2}{10} \cdot \frac{65}{100}}=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}} \\ & =\frac{160 / 1000}{510 / 1000}=\frac{16}{51}=0.313725=0.314 \end{aligned} \end{aligned}$$
A letter is known to have come either from "TATA NAGAR' or from 'CALCUTTA'. On the envelope, just two consecutive letters TA are visible. What is the probability that the letter came from 'TATA NAGAR'?
Let $E_1$ be the event that letter is from TATA NAGAR and $E_2$ be the event that letter is from CALCUTTA.
Also, let $E_3$ be the event that on the letter, two consecutive letters TA are visible.
$$\begin{array}{l} \therefore & P\left(E_1\right) =\frac{1}{2} \text { and } P\left(E_2\right)=\frac{1}{2} \\ \text { and } & P\left(E_3 / E_1\right)=\frac{2}{8} \text { and } P\left(E_3 / E_2\right)=\frac{1}{7} \end{array}$$
[since, if letter is from TATA NAGAR, we see that the events of two consecutive letters visible are $\{T A, A T, T A, A N, N A, A G, G A, A R\}$. So, $P\left(E_3 / E_1\right)=\frac{2}{8}$ and if letter is from CALCUTTA, we see that the events of two consecutive letters to visible are $\{C A, A L, L C, C U, U T, T, T A\}$.
So, $\left.P\left(E_3 / E_2\right)=\frac{1}{7}\right]$
$$\begin{aligned} & \therefore \quad P\left(E_1 / E_3\right)=\frac{P\left(E_1\right) \cdot P\left(E_3 / E_1\right)}{P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right)} \\ &=\frac{\frac{1}{2} \cdot \frac{2}{8}}{\frac{1}{2} \cdot \frac{2}{8}+\frac{1}{2} \cdot \frac{1}{7}}=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{14}}=\frac{1 / 8}{\frac{22}{8 \times 14}}=\frac{\frac{1}{8}}{\frac{11}{56}}=\frac{7}{11} \end{aligned}$$
There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3 , a ball is taken from the Ist bag but it shows up any other number, a ball is chosen from the II bag. Find the probability of choosing a black ball.
Since, Bag I $=\{3$ black, 4 white balls $\}$, Bag II $=\{4$ black, 3 white balls $\}$
Let $E_1$ be the event that bag $I$ is selected and $E_2$ be the event that bag $I I$ is selected.
Let $E_3$ be the event that black ball is chosen.
$\therefore \quad P\left(E_1\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$ and $P\left(E_2\right)=1-\frac{1}{3}=\frac{2}{3}$
$$\begin{array}{l} &\text { and } & P\left(E_3 / E_1\right) =\frac{3}{7} \text { and } P\left(E_3 / E_2\right)=\frac{4}{7} \\ & \therefore & P\left(E_3\right) =P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right) \\ & & =\frac{1}{3} \cdot \frac{3}{7}+\frac{2}{3} \cdot \frac{4}{7}=\frac{11}{21} \end{array}$$
There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.
$$\begin{array}{ll} \text { Let } & U_1=\{2 \text { white, } 3 \text { black balls }\} \\ & U_2=\{3 \text { white, } 2 \text { black balls }\} \\ \text { and } & U_3=\{4 \text { white, } 1 \text { black balls }\} \\ \therefore & P\left(U_1\right)=P\left(U_2\right)=P\left(U_3\right)=\frac{1}{3} \end{array}$$
Let $E_1$ be the event that a ball is chosen from urn $U_1, E_2$ be the event that a ball is chosen from urn $U_2$ and $E_3$ be the event that a ball is chosen from urn $U_3$.
Also, $$P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=1 / 3$$
Now, let $E$ be the event that white ball is drawn.
$$\therefore \quad P\left(E / E_1\right)=\frac{2}{5}, P\left(E / E_2\right)=\frac{3}{5}, P\left(E / E_3\right)=\frac{4}{5}$$
$$\begin{aligned} &\text { Now, }\\ &\begin{aligned} P\left(E_2 / E\right) & =\frac{P\left(E_2\right) \cdot P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right)} \\ & =\frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{1}{3} \cdot \frac{2}{5}+\frac{1}{3} \cdot \frac{3}{5}+\frac{1}{3} \cdot \frac{4}{5}} \\ & =\frac{\frac{3}{15}}{\frac{2}{15}+\frac{3}{15}+\frac{4}{15}}=\frac{3}{9}=\frac{1}{3} \end{aligned} \end{aligned}$$