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In the newspaper "Pop World," an old photo of well-known pop singers was published from when they were kids.

The names of the singers are known, and one needs to identify the singers. If the match is done randomly, what is the probability of identifying at least one of the first 2 singers?

So it is like ordering 4 people in a row to create a 1-1 and onto function between the names and the people, so $|\Omega|=4!$.

Now let $A,B,C,D$ be the event of guessing the name right, so we are looking for $(A\cup B)\cap(C^{c}\cap D^{c})$, so using the complement would not help? as it $(A^{c}\cap B^{c})\cup(C\cup D)$?

That is why we need to to calculate $P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{3!}{4!}+\frac{3!}{4!}-P(A\cap B)$. What should $P(A\cap B)$ be? Intuitively, it is $2!\over 4!$, but why isn't it $4*3\over 4!$?

(Maybe I should look at it this way: knowing the names of 2 people is equivalent to ordering the 2 that are left, so I have $2!$ ways of doing it. So that's why it is $2!\over 4!$, but still why is it $1*1*2*1$? Why does knowing the first and second remove higher numbers?)

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  • $\begingroup$ Are you allowed to guess the same name twice? $\endgroup$ Jul 6, 2015 at 10:07
  • $\begingroup$ @mathreadler the have 4 different name (I will add it, sorry) $\endgroup$
    – gbox
    Jul 6, 2015 at 10:10
  • $\begingroup$ Yes they got different names but if the objective is to only get one right you would have 100% to accomplish that if you guessed the same name on everyone. $\endgroup$ Jul 6, 2015 at 10:12
  • $\begingroup$ @mathreadler that can not be done, you must match one name to one person $\endgroup$
    – gbox
    Jul 6, 2015 at 10:14
  • $\begingroup$ But then order doesn't matter. What does "the two first" people mean? Getting at least one right out of two particular people? Or any two people? $\endgroup$ Jul 6, 2015 at 10:22

2 Answers 2

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You almost solved the problem yourself:

$\begin{align} P(A\cup B) &= P(A)+P(B)-P(A\cap B)\\ P(A)=P(B)&= \frac{1}{4}\\ P(A\cap B)&=P(A)P(B\mid A)\\ &=\frac{1}{4}\frac{1}{3}=\frac{2!}{4!}\\ P(A\cup B)&=\frac{5}{12} \end{align}$

The reason why $P(B\mid A)=\frac{1}{3}$ is simply that name used for the first guy is clearly not the one of the second one, since the first guess was right.

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Let the singers be (AB)(CD), we are interested in the first group

We need P(both right) + P(only 1st right) + P(1st from wrong group)*P(2nd right)

= $(\frac{1}{4}\cdot \frac{1}{3}) + (\frac{1}{4}\cdot\frac{2}{3}) + (\frac{1}{2}\cdot\frac{1}{3}) = \frac{5}{12}$

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    $\begingroup$ This is a good alternative approach. The two answers will come out the same, of course, because $P(A \cap B^\complement) = P(A) - P(A \cap B)$ and $P(A^\complement \cap B) = P(B) - P(A \cap B)$. $\endgroup$
    – David K
    Jul 6, 2015 at 12:03
  • $\begingroup$ @DavidK to which two answers the come the same you mean? $\endgroup$
    – gbox
    Jul 6, 2015 at 12:13
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    $\begingroup$ @gbox I mean this answer calculates $P(A\cap B) + P(A\cap B^\complement) + P(A^\complement \cap B)$, and this will always come to the same value as $P(A)+P(B)-P(A\cap B)$ because you can show those two formulas are equal. $\endgroup$
    – David K
    Jul 6, 2015 at 12:24
  • $\begingroup$ what does P(1st from wrong group)*P(2nd right) mean? $\endgroup$
    – gbox
    Jul 7, 2015 at 7:46
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    $\begingroup$ We took the singers to be (AB)(CD). What the sentence means is P[1st guess is from (CD)] ∩ P[2nd guess is B] $\endgroup$ Jul 7, 2015 at 7:53

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