If 38% of computer science students in north america last year were women, (a) what is the probability that the top mark was earned by a woman? (b) what is the probability that exactly 6 of the 10 top marks were earned by women? (c) how many women would you expect in the top 10?

A) The answer is 38%.

The event has only two possible outcomes: the top mark is earned either by a man or by a woman.
The events are also independent of each other: the top mark earner does not depend on who earned the second highest mark.
Therefore, we are talking about a binomial distribution, in which the probability of success (mark earned by a woman) is p(W) = 0.38, which means 38%.

B) The probability that exactly 6 of the 10 top marks were earned by women is 9.34%.

The probability of getting exactly 6 women in 10 marks is given by the formula:


where:
n = total number of events = 10
k = number of success we want = 6
p = probability of a succesfull event = 0.38

Substituting the numbers:


P(W = 6) = 210 · 0.00301 · 0.14776
              = 0.0934 

Hence, the probability of having 6 women earning among the top 10 marks is 0.0934, which means 9.34%

C) We would expect to have 3 women in the top 10.

We would expect that the percentage of the total population is the same of the top 10 marks, therefore:
W = n · p
    = 10 · 0.38
    = 3.8

Since we cannot have decimals of a physical person, the closest integer is 3.

It is 198,87032,750 + 375 = 33145 so, we take the 6% and we make it a decimal (0.6) and multiply that, then that is your answer