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Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls respectively. 1 of the bags is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the third bag.

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16

1118

913

415

answer is D.

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Detailed Solution

Given that there are 3 bags with balls of different colors.
First bag has 3 red and 7 black balls.
Second bag has 8 red and 2 black balls.
Third bag has 4 red and 6 black balls.
Let

E 1, E 2, E 3

and A are the events defined as follows.

E 1

= First bag is chosen

E 2

= Second bag is chosen

E 3

= Third bag is chosen
A = Ball drawn is red
The probability of an event happening is

\frac{Number of favorable outcomes}{Total number of outcomes}

.
Since there are three bags and one of the bags is chosen at random, so,

P E 1 = P E 2 = P E 3 = 13

.
If

E 1

has already occurred, then the first bag has been chosen which contains 3 red and 7 black balls.
The probability of drawing 1 red ball from it is

310

.
So,

P A / E 1 = 310

.
Now, if

E 2

has already occurred, then a second bag has been chosen which contains 8 red and 2 black balls.
The probability of drawing 1 red ball from it is

810

.
So,

P A / E 2 = 810

.
Now, if

E 3

has already occurred, then a second bag has been chosen which has 4 red and 6 black balls.
The probability of drawing 1 red ball from it is

410

.
So,

P A / E 3 = 410

.
Therefore, the required probability is,

= P E 3 × P A / E 3 P E 1 × P A / E 1 + P E 2 × P A / E 2 + P E 3 × P A / E 3

= 13 × 410 13 × 310 + 13 × 810 + 13 × 410

= 430 330 + 830 + 430

= 4301530

= 415

The probability that a red ball is chosen from the third bag is

\frac{4}{15} .

Hence, option 4 is correct.

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