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Q.

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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a

  1 6

b

11 18

c

9 13

d

4 15  

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 Number of favorable outcomesTotal 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 = 1 3 .
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 3 10 .
So, P A/ E 1 = 3 10 .
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 8 10 .
So, P A/ E 2 = 8 10 .
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 4 10 .
So, P A/ E 3 = 4 10 .
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
= 1 3 × 4 10 1 3 × 3 10 + 1 3 × 8 10 + 1 3 × 4 10
= 4 30 3 30 + 8 30 + 4 30
= 4 30 15 30
= 4 15
The probability that a red ball is chosen from the third bag is 415.
Hence, option 4 is correct.
 
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