A basket contains 5 RED and 3 BLACK chips You randomly pick

A basket contains 5 RED and 3 BLACK chips. You randomly pick a chip from the basket, and if it is RED chip, you keep that chip outside and pick another chip. On the other hand, if you pick a BLACK chip, you put the chip back to the basket along with a new RED chip. Assume you have just picked the second chip now and answer the following questions

What is the probability that both chips picked are RED?

If the second picked chip is RED, what is the probability that first picked chip was BLACK?

Solution

1.

The probability of the first chip being red is 5/(3 + 5) = 5/8.

P(1st is red) = 5/8.

Now, as it is red, you put it outside, so only 4 out of the 7 remaining is red.

Thus,

P(2nd is red|1st is red) = 4/7

Thus,

P(both chips are red) = (5/8)(4/7) = 5/14 [ANSWER]

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2.

By Bayes\' Rule,

P(2nd is red) = P(1st is red) P(2nd is red|1st is red) + P(1st is black) P(2nd is red|1st is black)

As

P(first is red) = 5/8
P(second is red|1st is red) = 4/7
P(1st is black) = 3/8
P(2nd is red|1st is black) = 6/9

Thus, plugging these in,

P(2nd is red) = 17/28

Thus,

P(1st is black|2nd is red) = P(1st is black and 2nd is red) / P(2nd is red)

As P(1st is black and 2nd is red) = P(1st is black) P(2nd is red|1st is black) = (3/8)(6/9) = 1/4

Then

P(1st is black|2nd is red) = (1/4) / (17/28) = 7/17 [ANSWER]