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A bag contains 4 white and 2 black balls and another bag contains 3 of each colour. A bag is selected at random and a ball is drawn at random from the bag chosen. The probability of the white ball drawn is
1. \(\frac{1}{3}\)
2. \(\frac{1}{4}\)
3. \(\frac{5}{{12}}\)
4. \(\frac{7}{{12}}\)

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Correct Answer - Option 4 : \(\frac{7}{{12}}\)

Concept:

Total Probability Theorem:

Let events C1, C2 . . . Cn form partitions of the sample space S, where all the events have a non-zero probability of occurrence. For any event, A associated with S, according to the total probability theorem,

\(P\left( A \right) = \mathop \sum \limits_{k = 0}^n P\left( {{C_k}} \right)P\left( {A/{C_k}} \right)\)

Calculation:

Bag 1: 4 White, 2 Black

Bag 2: 3 White, 3 Black

Let P(B1) is the probability of selecting bag 1

P(B2) is the probability of selecting bag 2

P(W) is the probability of selecting white

P(W/B1) is the probability of selecting white given that the selected bag is Bag 1

P(W/B2) is the probability of selecting white given that the selected bag is Bag 2

As there are two bags, the probability selecting any of the bag is same and it is equal to

\(P\left( {B1} \right) = P\left( {B2} \right) = \frac{1}{2}\)

If the selected bag is Bag1, the probability of selecting white is

As there are 4 whites out of total 6 balls

\(P\left( {W/B1} \right) = \frac{4}{6} = \frac{2}{3}\)

If the selected bag is Bag2, the probability of selecting white is

As there are 3 whites out of total 6 balls

\(P\left( {W/B2} \right) = \frac{3}{6} = \frac{1}{2}\)

Now, from the total probability theorem

\(P\left( W \right) = P\left( {W/B1} \right)P\left( {B1} \right) + P\left( {W/B2} \right)P\left( {B2} \right)\)

\( = \frac{2}{3} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{7}{{12}}\)

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