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I am self-learning basic undergrad calculus-based probability. I would like someone to verify if my solution is correct, especially part(b). I am posting my attempt/solution below:

[BH 4.23] Alice and Bob have just met and wonder, whether they have a mutual friend. Each has $50$ friends out of $1000$ other people who live in their town. They think that its unlikely that they have a friend in common, saying each of us is only friends with $5$ percent of the people here, so it would be very unlikely that our two $5$ percents overlap.

Assume that Alice's $50$ friends are a random sample of the $1000$ people (equally likely to be any $50$ of the $1000$), and similarly for Bob. Also, assume that knowing who Alice's friends are gives no information about who Bob's friends are.

a) Compute the expected number of mutual friends Alice and Bob have.

b) Let $X$ be the number of mutual friends they have. Find the PMF of $X$.

c) Is the distribution of $X$ one of the important distributions we have looked at? If so which?

Solution. (My Attempt)

a) Let $A_k$ be the event that person-$k$ is Alice's friend.

Let $B_k$ be the event that person-$k$ is Bob's friend.

Let $I_{A_k}$ be the indicator function of the event $A_k$, let $I_{B_k}$ be the indicator function of the event $B_k$.

Since $I_{A_k}$ and $I_{B_k}$ are independent, $E(I_{A_k} \cdot I_{B_k}) = P(A_k,B_k) = P(A_k)\cdot P(B_k)=E(I_{A_k})E(I_{B_k})$.

Let $X$ be the number of mutual friends of Alice and Bob. Then,

\begin{align*} X &= \sum_{k=1}^{1000}I_{A_k}\cdot I_{B_k}\\ E(X) &= 1000 \times 0.05 \times 0.05 = 2.5 \end{align*}

b) $X$ can take values in the set $\{0,1,2,\ldots,50\}$.

The probability that a pre-specified friend of Alice is also a friend of Bob is given by,

\begin{align*} P(B_k|A_k) = P(B_k) \end{align*}

We are restricting our population to Alice's $50$ friends, since the frequentist interpretation of conditional probabilities is,

\begin{align*} P \{B_k | A_k\} = \frac{P \{B_k, A_k\}}{P\{A_k\}} = \frac{n(B_k,A_k)}{n(A_k)} \end{align*}

Each of Alice's friends have a $p=0.05$ chance of being a friend of Bob. Further, the occurrence of one mutual friend does not reveal any information about the occurrence of another mutual friend. These are independent and identical Bernoulli trials with success probability $p=0.05$ .

So, the PMF of $X$ is:

\begin{align*} P(X = k) = {50 \choose k} (0.05)^k (0.95)^{50 - k} \end{align*}

Note, that $E(X) = np = 50 \times 0.05 = 2.5$ which ties up with the previous result.

c) $X \sim Bin\left(n=50,p=\frac{1}{20}\right)$

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This part "the occurrence of one mutual friend does not reveal any information about the occurrence of another mutual friend" is not true.

I think it is more straightforward to treat the event $\{X=k\}$ as "k mutual friends, $50-k$ only friends for Alice and $50-k$ only friends for Bob". Then we have $$ P(X=k)=\frac{{1000 \choose k}{1000-k \choose 50-k}{950 \choose 50-k}} {{1000 \choose 50}^2} $$ And after calculation we can find this is Hypergeometric distribution($n=50,N=1000,k=50$).

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  • $\begingroup$ Hi, suppose that we think of Alice's friends as white balls and non-friends of Alice as black balls. $w = 50$, $b=950$. We want construct a sample of size $50$(friends of Bob) without replacement, such that it contains exactly $k$ balls. The number of such distinguishable samples are ${50 \choose k}{950 \choose 50 - k}$. Since each of the ${1000 \choose 50}$ samples are equally likely, the probability that it contains $k$ white balls is $\frac{{50 \choose k}{950 \choose 50 - k}}{{1000 \choose 50}}$. Does this approach work? $\endgroup$
    – Quasar
    Dec 31, 2021 at 12:01
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    $\begingroup$ @Quasar Yes, that is correct. $\endgroup$
    – ago yang
    Dec 31, 2021 at 14:08

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