Question

6. Someone has a total of 10 pens. Of them, 6 are black and 4 others are green in color. He picks randomly a pen one day and then grabs another pen next day without replacing the first one. The probability of the first pen to be black is P(A) = 6/10 .

a) What is the probability that the second pen grabbed is black given that the first pen is also black, P(BlA) = 5/9 .

b) Now, what is the probability that both pens are black? P(A and B) = P(A)P(BlA) = (6/10) (5/9).

Answer 1

Req A:

Total Pens: 10

Black pen: 6

Green pen: 4

The probability of getting the black pen in first draw p (A): 6 /10

Now, after the first pen drawn as black (and this has not been replaced while drawing second pen), therefore,

During the second draw:

Black pen left: 6-1 = 5

Total pen: 10-1= 9

Therefore, probability of getting the second pen as balck p(b/A) = 5/9

Req B: The probability of getting both the pens black = Probability of first pen balck in first draw p(A) and Probability of getting blanck pen in second draw p(B/A) = p (A) * p(B/A) = 6/10 * 5/9 = 30/90= 1/30