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  • 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 s

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 second bag. Find the probability of choosing a black ball.

Answers (1)

Solution

Given-

There are 2 bags-

Bag 1: 3 black and 4 white balls
Bag 2: 4 black and 3 white balls

Therefore, Total balls = 7

Let events E1, E2 be the following:

E1  and E2 be the events that bag 1 and bag 2 are selected respectively

We know that a die is thrown.

Therefore, total outcomes = 6

\therefore P\left(E_{1}\right)=\frac{2}{6}=\frac{1}{3} \text { and } P\left(E_{2}\right)=1-\frac{1}{3}=\frac{2}{3}

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
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)

Let “E” be the event that the black ball is chosen.
P(E|E1) is the probability that a black ball is chosen from the bag 1.
P(E|E2) is the probability that a black ball is chosen from the bag 2.
Therefore,

\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{3}{7}$ and $\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=\frac{4}{7}

Therefore, the probability of choosing a black ball is the sum of individual probabilities of choosing the black from the given bags.

From the law of total probability,
\\\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right) \\=\frac{1}{3} \times \frac{3}{7}+\frac{2}{3} \times \frac{4}{7} \\=\frac{3}{21}+\frac{8}{21}=\frac{11}{21}

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infoexpert22

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