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I am a guy (x) who knows four other guys (y,z,v,w) and five women (a,b,c,d,e). On a given day, we all speak to another person with probability $p$. On a given day, how many girls I either speak with directly or I spoke with someone (boy or girl) who also spoke to her?

This is what I have done. I spoke with 2.5 women in expectation and 2 guys. I do not know how to add the women those guys spoke with that I did not, and the women that the women I spoke with spoke with, adjusting for the fact that they could be the same.

Okay tried again: if women do not know each other, the answer is that I know one woman with probability $p$+ I do not know her $1-p$, so

 $p+p(1-q)+p(1-q)^2+p(1-p)^3+p(1-p)^4$

yet this answer is not completely good because women know each other with probability $p$.

fox
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  • The first part should be pretty easy. You know what the probability of talking to a woman is, so what is the expected number of women that you speak with? – Matti P. May 09 '18 at 11:41
  • I think you need something along $P(A \dot\cup B) = P( A \cup B ) - P (A \cap B)$. –  May 09 '18 at 11:49

1 Answers1

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Let $A$ be the event that either you speak to $a$ or you speak to someone who speaks to $a$, and let $A'$ be the complement of $A$ (i.e., $A' = \text{not}\,A$).

Let $k$ be the number of people other than $a$ that you speak to.

Let $q=1-p$. \begin{align*} \text{Then}\;\;P(A') &=q\left(\sum_{k=0}^8\left(\binom{8}{k}p^kq^{8-k}\right)q^k\right)\\[4pt] &=q\left(\sum_{k=0}^8\binom{8}{k}p^kq^8\right)\\[4pt] &=q^9\!\left(\sum_{k=0}^8 \binom{8}{k}p^k\right)\\[4pt] &=q^9\left(1+p\right)^8\\[4pt] \end{align*} Noting that the contribution of $A$ to the required expected value is just $P(A)$, it follows that the required expected value is just $$5P(A) = 5(1-P(A'))=5\left(1-\left(q^9\left(1+p\right)^8\right)\right)$$

More generally, to allow for easy testing, suppose there are $m$ men (including you), and $w$ women.

Then letting $n=m+w$, the expected value would be $$w\left(1-\left(q^{n-1}\left(1+p\right)^{n-2}\right)\right)$$ Testing the above formula using $$m=2,\;w=1,\;p={\small{\frac{1}{2}}}$$ yields an expected value of ${\large{\frac{5}{8}}}$, which can easily be verified by direct calculation.

quasi
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    @Josué Ortega: Analyze the first line in the derivation of $P(A')$, The leading factor $q$ expresses the condition that you didn't speak to $a$. Inside the parentheses, we have cases, based on the number $k$ of people other than $a$ that you spoke to, First you have to choose them, which explains the factor $C(8,k)$. Then you have to speak to all $k$ of them, and not speak to the $8-k$ others, which explains the factor $p^kq^{8-k}$. Then for all of the $k$ others that you spoke to, we need for them not to speak to $a$, which explains the factor $q^k$. The lines after are just simplifications. – quasi May 09 '18 at 15:39
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    @Josué Ortega: That's the Binomial Theorem. – quasi May 10 '18 at 11:48
  • @Josué Ortega: Connected by a path of length at most 2? Also, for the new question, what does gender have to do with it? – quasi May 10 '18 at 15:56
  • @Josué Ortega: It appears to have been answered several days ago. Is that answer not sufficient? – quasi May 10 '18 at 16:47
  • I do not know, I understood your answer to this thread very well, but I did not understand that one. I tried generalizing your answer so $P(A')=P($I do not know someone from my group$)=p \sum \sum p^k q^h (1-p)^{n-k}(1-q)^{n-h} (1-p)^{k}(1-q)^{h}$, but I am missing the combinatorial n choose k analogous – fox May 10 '18 at 16:51
  • @Josué Ortega: I'm on my way out. If there's no new accepted answer to the new question when I get back (in about $4$ hours), I'll look at it then. – quasi May 10 '18 at 16:57