Question

What if you had 45 males and 55 females in a class and you put their names on separate pieces of paper and placed them in a bowl. That would be the probability of selecting a male in the first draw, a female in the second draw and a female in the third draw? You do not replace names after you drawn them out of the bowl. Round your answer to the nearest 3 decimal places.

Answer 1

Before first draw there are 45 male names and 55 female names.

Let A be the event of selecting a male in the first draw, B be the event of selecting a female in the second draw and C be the event of selecting a female in the third draw.

After drawing a male name in the first draw, there are 44 male names and 55 female names.

After drawing a female in the second draw, there are 44 male and 54 female names.

So, we have, P(A) = 45/(45 + 55) = 45/100 = 0.45

P(B | A) = 55/(44 + 55) = 55/99 = 0.5556

P(C | B A) = 54/(44 + 54) = 54/98 = 0.5510

The required probability = P(A B C) = P(A) P(B | A) P(C | B A) = 0.45 * 0.5556 * 0.5510 = 0.1378 (Ans).