Multiple choice

There are two bags, one of which contains $3$ black and $4$ white balls,$II$ bag contain $4$ black and $3$ white balls. A die is cast, if the number $3$ or less than $3$ turns up, a ball is drawn from $1$ bag and if the face greater than $3$ turns up, a ball is drawn from $II$ bag. It is found that ball drawn is black. Find the probability that it is from bag I.

  1. $\dfrac {3}{14}$
  2. $\dfrac {4}{7}$
  3. $\dfrac {2}{7}$
  4. $\dfrac {3}{7}$
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D Correct answer
Explanation

P(Bag I) = 1/2, P(Black|Bag I) = 3/7. P(Bag II) = 1/2, P(Black|Bag II) = 4/7. P(Bag I|Black) = (P(B|I)P(I)) / (P(B|I)P(I) + P(B|II)P(II)) = (3/7 * 1/2) / (3/7 * 1/2 + 4/7 * 1/2) = 3/7.

AI explanation

Using Bayes theorem, we find the probability that the black ball came from bag I. P(Bag I) is 3/6, P(Bag II) is 3/6, P(Black from I) is 3/7, and P(Black from II) is 4/7. The calculation is ((1/2) * (3/7)) / (((1/2) * (3/7)) + ((1/2) * (4/7))), which evaluates to (3/14) / (7/14). The result is 3/7.

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