Probability stranger in a group

Question

Given a group of n = 100 people made up of Italians and foreigners, two people are chosen at random. Knowing that the probability that only one of them is foreign is p ≈ 0.18, a possible composition of the group is identified (number of Italians nI and number of foreigners nS).

I wrote that $ \binom{100}{1} p_s ^{1} ( 1- p_s)^{99} \binom{99}{1} p_s^{0} {1-p_s }^{99} = 0.18 $ so i found that $ p_s = 0.053 $ where $ p_s $ is the probability that a person is stranger. I dint know if a made mistakes because I dint have the result

Answer 1

If I understand your problem correctly, you should just find ways to choose $2$ people out of $100$ such that one of them is foreigner (and the other is Italian).

Say there are $n$ foreigners. Then we have,

$\displaystyle \frac{{n \choose 1} {100-n \choose 1}}{100 \choose 2} = 0.18$

Leads to quadratic $n^2 - 100n + 891 = 0$. You can solve for possible values taking nearest positive integers.

Answer 2

Let $n_I$ be the number of Italians, and $n_S$ be the number of foreigners. We know that $$n_I + n_S = n = 100.$$

We are told that if we choose two (presumably different) people at random from the group, the probability that exactly one of the two is foreign is $p = 0.18$. We count the number of ways to choose two distinct people from the group such that one is foreign and one is Italian. There are $n_I$ ways to choose one person from the group of Italians; and $n_S$ ways to choose one person from the group of foreigners. Since these choices may be made independently without restriction, there are a total of $n_I n_S$ ways to select two people from the whole group so that exactly one is foreign. But there are $$\binom{n}{2} = \frac{n(n-1)}{2}$$ ways to choose any two people from the group of $100$. So the given condition is equivalent to $$\frac{n_I n_S}{n(n-1)/2} = p = 0.18.$$ We now substitute $n = 100$ and $n_S = 100 - n_I$ to obtain $$n_I(100 - n_I) = 4950(0.18) = 891,$$ and solve the resulting quadratic for $n_I$: $$n_I = 50 \pm \sqrt{1609} \approx \{9.88766, 90.1123\}.$$ So rounding to the nearest integer gives us the solution set $$n_I \in \{10, 90\}.$$ Notice that both solutions are valid--it is impossible to determine with the information given whether the group comprises $90$ Italians and $10$ foreigners, or $10$ Italians and $90$ foreigners. This should be obvious because the roles of "Italians" and "foreigners" may be interchanged and the question would remain identical, since the outcome of one foreigner among two people selected is the same as saying there is one Italian among two people selected.

Since we rounded the result, this means that the precise value of $p$ is no longer $0.18$. If we use $n_I \in \{ 10, 90\}$, what is the resulting value of $p$?

Answer 3

Elena - in your “binomial speak”, we have two “experiments” and we are looking at probability of seeing one success i.e. one foreigner.

$\binom{2}{1} p_s (1-p_s) = 0.18$