dabaobao wrote:
There is a group of people consisting of 10 men and 6 women. Among these 16 people, there are 4 married couples (manwoman couples). If one man and one woman are selected at random, what is the probability that a married couple gets selected?
A) 1/30
B) 1/15
C) 7/30
D) 1/4
E) 1/2
Veritas Prep Official Solution
P(Selecting a Married Man) = 4/10
P(Selecting the Wife of that Man) = 1/6
P(Married Couple is Selected) = (4/10)*(1/6) = 4/60 = 1/15
Explanation
We need to select a couple. Here, it doesn’t matter whether you select the man first or the woman; the two member types are different and there is only one way in which you can select the corresponding partner. You cannot select two members of the same type e.g. two men or two women. Hence we don’t need to bother with calculating the different cases of selecting the man first or the woman first.
Of course, even if we do it, we will get the correct answer. Let me show you the calculation.
The total number of ways of selecting a man and a woman are ‘select a man in 10 ways’ and ‘a woman in 6 ways’. Then ‘select a woman in 6 ways’ and ‘then a man in 10 ways’ i.e. total 120 ways. To select a couple, you can select a man in 4 ways and the woman in 1 way. You can select a woman in 4 ways and the man in 1 way. So total 4 + 4 = 8 ways.
Probability of selecting a couple = 8/120 = 4/60 (same as before).