Three bags contain a number of red and white balls as follows Bag I: 3 red balls, Bag II : 2 red balls and 1 white ball and Bag III : 3 white balls. The probability that bag i will be chosen and a ball is selected from it is $\frac{i}{6}$, where $i=1,2,3$. What is the probability that
(i) a red ball will be selected?
(ii) a white ball is selected?
Bag I: 3 red balls and 0 white ball.
Bag II : 2 red balls and 1 white ball.
Beg III : 0 red ball and 3 white balls.
Let $E_1, E_2$ and $E_3$ be the events that bag I , bag II and bag III is selected and a ball is chosen from it.
$$P\left(E_1\right)=\frac{1}{6}, P\left(E_2\right)=\frac{2}{6} \text { and } P\left(E_3\right)=\frac{3}{6}$$
(i) Let $E$ be the event that a red ball is selected. Then, probability that red ball will be selected
$$\begin{aligned} P(E) & =P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right) \\ & =\frac{1}{6} \cdot \frac{3}{3}+\frac{2}{6} \cdot \frac{2}{3}+\frac{3}{6} \cdot 0 \\ & =\frac{1}{6}+\frac{2}{9}+0 \\ & =\frac{3+4}{18}=\frac{7}{18} \end{aligned}$$
$$\begin{aligned} &\text { (ii) Let } F \text { be the event that a white ball is selected. }\\ &\begin{aligned} & \therefore \quad \begin{aligned} P(F) & =P\left(E_1\right) \cdot P\left(F / E_1\right)+P\left(E_2\right) \cdot P\left(F / E_2\right)+P\left(E_3\right) \cdot P\left(F / E_3\right) \\ & =\frac{1}{6} \cdot 0+\frac{2}{6} \cdot \frac{1}{3}+\frac{3}{6} \cdot 1=\frac{1}{9}+\frac{3}{6}=\frac{11}{18} \end{aligned} \\ & \text { Note } P(F)=1\left.-P(E)=1-\frac{7}{18}=\frac{11}{18} \text { [since, we know that } P(E)+P(F)=1\right] \end{aligned} \end{aligned}$$
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 $P\left(E_2\right)=1-\frac{1}{3}=\frac{2}{3}$
$$\begin{array}{l} &\text { and } & P\left(E_3 / E_1\right) =\frac{3}{7} \text { and } P\left(E_3 / E_2\right)=\frac{4}{7} \\ & \therefore & P\left(E_3\right) =P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right) \\ & & =\frac{1}{3} \cdot \frac{3}{7}+\frac{2}{3} \cdot \frac{4}{7}=\frac{11}{21} \end{array}$$