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GO Classes DPP | Engineering Mathematics | Probability | JEE Mains

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Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively, One of the um is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn $A$ is:

  1. $\frac{4}{17}$
     
  2. $\frac{5}{16}$
     
  3. $\frac{5}{18}$
     
  4. $\frac{7}{18}$

5 Answers

0 0 votes
  1. Explanation : Ans - c

    • Urn A: 7 red balls, 5 black balls (total 12 balls).

    • Urn B: 5 red balls, 7 black balls (total 12 balls).

    • Urn C: 6 red balls, 6 black balls (total 12 balls).

    • One urn is selected at random with equal probability, so P(A)=P(B)=P(C)=1/3​.

    • A ball is drawn from the selected urn.

  2. Total Number of Black Balls:

    • Number of black balls in urn A: 5C1(since we are choosing 1 black ball from 5, but this is equivalent to counting 5 black balls).

    • Number of black balls in urn B: 7C1 =7.

    • Number of black balls in urn C: 6C1=6.

    • Total black balls across all urns: 5+7+6=18.

  3.  # of favourable outcomes =  # of black ball in UrnA = 5C1 =5

 Finally prob (urnA | black ball ) = # of favourable outcomes( # of black ball in urnA) / total number of black ball .

                                                     = 5/18 

 

Note : Prob (black ball) = Prob( urnA ∩ black ball ) + Prob(UrnB ∩ black ball) + Prob( UrnC ∩ black ball)

0 0 votes

P(A) = P(B) = P(C) = 1/3

P(A | B) = P(B | A) x P(A) / P(B) 
             = (5/12) x (1/3) / (18/36)
             = (5/36) x (2/1)
            = 5/18..

So, OPTION C is correct..!!

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