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  • Three bags contain a number of red and white balls as follows: 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, i = 1, 2, 3. What is the probab

Three bags contain a number of red and white balls as follows:
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, i = 1, 2, 3. What is the probability that
(i) a red ball will be selected? (ii) a white ball is selected?

Answers (1)

Let E1, E2, and E3 be the events that Bag 1, Bag 2 and Bag 3 are 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.

\\ \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{6}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{2}{6}$ and \\$\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{3}{6}$
The Law of Total Probability:
In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
$\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)+\ldots \ldots \mathrm{P}\left(\mathrm{E}_{n}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{n}\right)$

(i) Let “E” be the event that a red ball is selected.
P(E|E1) is the probability that red ball is chosen from the bag 1.
P(E|E2) is the probability that red ball is chosen from the bag 2.
P(E|E3) is the probability that red ball is chosen from the bag 3.

Therefore,

\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{3}{3}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=\frac{2}{3}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{3}\right)=0

As red ball can be selected from Bag 1, Bag 2 and Bag 3.

Therefore, probability of choosing a red ball is the sum of individual probabilities of choosing the red from the given bags.
From the law of total probability,

\\ P(E)=P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(E \mid E_{2}\right)+P\left(E_{3}\right) \times P\left(E \mid E_{3}\right) \\ =\frac{1}{6} \times \frac{3}{3}+\frac{2}{6} \times \frac{2}{3}+\frac{3}{6} \times 0 \\ =\frac{1}{6}+\frac{2}{9} \\ =\frac{3+4}{18} \\ =\frac{7}{18}

Let F be the event that a white ball is selected.

Therefore, P(F|E1) is the probability that white ball is chosen from the bag 1.

P(F|E2) is the probability that white ball is chosen from the bag 2.
P(F|E3) is the probability that white ball is chosen from the bag 2.
P(F|E1) = 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$
As white ball can be selected from Bag 1, Bag 2 and Bag 3
Therefore, sum of individual probabilities of choosing the red from the given bags is the probability of choosing a white ball.
\\ \mathrm{P}(\mathrm{F})=\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{1}{6} \times 0+\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times 1$ \\$=\frac{2}{18}+\frac{3}{6}$ \\$=\frac{2+9}{18}$ \\$=\frac{11}{18}$

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