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  • Refer to Question 41 above. If a white ball is selected, what is the probability that it came from (i) Bag 2 (ii) Bag 3

Refer to Question 41 above. If a white ball is selected, what is the probability that it came from
(i) Bag 2 (ii) Bag 3

Answers (1)

Referring to the previous question, using Bayes theorem, we get

Let E1, E2, and E3 be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.

Bag 1: 3 red balls,
Bag 2: 2 red balls and 1 white ball
Bag 3: 3 white balls.

The probability that bag i will be chosen and a ball is selected from it is i|6.

\\ P\left(E_{1}\right)=\frac{1}{6}, P\left(E_{2}\right)=\frac{2}{6}$ and \\$P\left(E_{3}\right)=\frac{3}{6}$
Let F be the event that a white ball is selected. Therefore, \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)$ is the probability that white ball is chosen from the bag 1 .
\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)$ is the probability that white ball is chosen from the bag 2 .
\\ \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)$ is the probability that white ball is chosen from the bag 2 . \\$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)=0$ \\$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)=\frac{1}{3}$ \\$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)=\frac{3}{3}=1$
To find: the probability that if white ball is selected, it is selected from:

(i) Bag 2
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

\begin{aligned} &\therefore \mathbf{P}(\mathbf{A} \mid \mathbf{B})=\frac{\mathbf{P}(\mathbf{A}) \mathbf{P}(\mathbf{B} \mid \mathbf{A})}{\mathrm{P}(\mathrm{B})}\\ &\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{F}\right) \text { is the probability that white ball is selected from bag } 2 \text { . }\\ &\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)}\\ &=\frac{\frac{2}{6} \times \frac{1}{3}}{\frac{1}{6} \times 0+\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times 1} \end{aligned}

$$ =\frac{\frac{1}{9}}{0+\frac{1}{9}+\frac{1}{2}} \\ =\frac{\frac{1}{9}}{\frac{2+9}{18}} \\ =\frac{1}{9} \times \frac{18}{11} \\ =\frac{2}{11} $$
(ii)Bag 3
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
\therefore \mathbf{P}(\mathbf{A} \mid \mathbf{B})=\frac{\mathbf{P}(\mathbf{A}) \mathbf{P}(\mathbf{B} \mid \mathbf{A})}{\mathrm{P}(\mathrm{B})}$

\mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{F}\right)$ is the probability that white ball is selected from bag 2

Using Bayes' theorem, we get the probability of \mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{F}\right)$ as:
\mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{3}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)}$ $=\frac{\frac{3}{6} \times 1}{\frac{1}{6} \times 0+\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times 1}$ $\\=\frac{\frac{1}{2}}{0+\frac{1}{9}+\frac{1}{2}}$

\\ =\frac{\frac{1}{2}}{\frac{2+9}{18}} \\ =\frac{1}{2} \times \frac{18}{11} \\ =\frac{9}{11}

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