Question

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?

Answer 1

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:

`P(X) = (n!)/(k!(n-k)!) p^(k)(1 - p)^(n-k)`

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) = (10!)/(6!(10-6)!) 0.38^(6)(1 - 0.38)^(10 - 6) \\ = (10!)/(6!(4)!) 0.38^(6)(0.62)^(4)`

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.