Two dice are tossed. Find whether the following two events $A$ and $B$ are independent $A=\{(x, y): x+y=11\}$ and $B=\{(x, y): x \neq 5\}$, where $(x, y)$ denotes a typical sample point.
$$\begin{aligned} & \text { We have, } A=\{(x, y): x+y=11\} \text { and } B=\{(x, y): x \neq 5\} \\ & \begin{aligned} \therefore & A=\{(5,6),(6,5)\}, B=\{(1,1),(1,2),(1,3),(1,4),(1,5)(1,6),(2,1),(2,2),(2,3),(2,4) \\ & (2,5)(2,6),(3,1),(3,2),(3,3),(3,4),(3,5)(3,6),(4,1),(4,2),(4,3),(4,4),(4,5)(4,6), \\ & (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\begin{array}{ll} \Rightarrow & n(A)=2, n(B)=30 \text { and } n(A \cap B)=1 \\ \therefore & P(A)=\frac{2}{36}=\frac{1}{18} \text { and } P(B)=\frac{30}{36}=\frac{5}{6} \\ \Rightarrow & P(A) \cdot P(B)=\frac{5}{108} \text { and } P(A \cap B)=\frac{1}{36} \neq P(A) \cdot P(B) \end{array}\\ &\text { So, } A \text { and } B \text { are not independent. } \end{aligned}$$
Q. 40 An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on $k$.
$$\begin{aligned} \text{Let }\quad U & =\{m \text { white, } n \text { black balls }\} \\ E_1 & =\{\text { First ball drawn of white colour }\} \\ E_2 & =\{\text { First ball drawn of black colour }\} \end{aligned}$$
$$\begin{aligned} &\text { and } \quad E_3=\{\text { Second ball drawn of white colour }\}\\ &\therefore \quad P\left(E_1\right)=\frac{m}{m+n} \text { and } P\left(E_2\right)=\frac{n}{m+n} \end{aligned}$$
$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} P\left(E_3 / E_1\right) & =\frac{m+k}{m+n+k} \text { and } P\left(E_3 / E_2\right)=\frac{m}{m+n+k} \\ \therefore\quad P\left(E_3\right) & =P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right) \\ & =\frac{m}{m+n} \cdot \frac{m+k}{m+n+k}+\frac{n}{m+n} \cdot \frac{m}{m+n+k} \\ & =\frac{m(m+k)+n m}{(m+n+k)(m+n)}=\frac{m^2+m k+n m}{(m+n+k)(m+n)} \\ & =\frac{m(m+k+n)}{(m+n+k)(m+n)}=\frac{m}{m+n} \end{aligned} \end{aligned}$$
Hence, the probability of drawing a white ball does not depend on k.
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]
$$\begin{aligned} &\text { (iii) }\\ &\begin{aligned} P\left(A_2 / \bar{E}\right) & =\frac{P\left(A_2\right) \cdot P\left(\bar{E} / A_2\right)}{P\left(A_1\right) \cdot P\left(\bar{E} / A_1\right)+P\left(A_2\right) \cdot P\left(\bar{E} / A_2\right)+P\left(A_3\right) \cdot P\left(\bar{E} / A_3\right)} \\ & =\frac{\frac{4}{10} \cdot \frac{40}{100}}{\frac{4}{10} \cdot \frac{55}{100}+\frac{4}{10} \cdot \frac{40}{100}+\frac{2}{10} \cdot \frac{65}{100}}=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}} \\ & =\frac{160 / 1000}{510 / 1000}=\frac{16}{51}=0.313725=0.314 \end{aligned} \end{aligned}$$