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**Example 1: **An urn contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

**Solution: **Let A = Getting a white ball in the first draw B = Getting a Black Ball in the second draw. Required Probability = Probability of getting a white ball in the first draw and a black ball in second draw = P(A and B) = P (A âˆ©B) = P (A) .P(B/A)

P(A) = (^{10}C_{1}/^{25}C_{1}) = (10/25) = (2/5) and P(B/A) = Probability of getting a black ball in the second draw

when a white ball has already been in first draw = (^{15}C_{1}/^{24}C_{1}) = (15/24) = (5/8)

â‡’ Required probability

If we know D has occurred, the possible outcomes are {1,2,3}.

For C to happen the outcome must be in Câˆ©D={1,3}.

P(C|D) = P(Câˆ©D)/P(D) = 2/3

**Example3:** Hunar wrote two sections of CAT paper ; Verbal and QA in the same order. The probability of her passing both sections is 0.6. The probability of her passing the verbal section is 0.8. What is the probability of her passing the QA section given that she has passed the Verbal section?

Now P(A/E1) = Probability of drawing a golden crystal when the first pouch has been chosen =3/7 and P(A/E2)=Probability of drawing an golden crystal when the second pouch has been selected = 2/5

[âˆ´ The second pouch contains 2 golden and 4 silver crystals]

Using the law of total probability P(golden crystal) = P(A) = P(E_{1}) .

Then by Law of total probability: P(M) = P (M| L) ^{.} P(L) + P (M | NL) ^{.} P(NL)

= (2/7).(1/4) + (1/7).(3/4) = 5/28

Then by the Law of total probability:

P(A) = P(A | X)-P(X) + P(A | Y) P(Y ) + P(A/Z).P(Z).

= (4/ 10).(1/ 3) + (1/ 6) . (1/ 3) + (3/8) .(1/3) = 113/360 .

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