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Three students are taking an exam independent of one another. Probability of first one passing the exam is 2/3, second one 3/4 and third one 4/5. If exactly one student passes the exam, what is the probability of it being the first student?

Here is my flow of thoughts:

A1 - first student passing
A2 - second student passing
A3 - third student passing
A - everyone has passed the exam

Since events in question are independent of one another,

P(A) = P(A1) * P(A2) * P(A3) =
= 0.66 * 0.75 * 0.80 = 0,396

So probability of everyone passing the exam is 0.396.

How do I get the probability of first one passing? Naive approach would be to divide it by 3, since there are three students but this must be somehow weighted since not everyone had the same probability of passing.

Is it as simple as finding LCM(3,4,5)=60 and multiplying by 20, since 60/3=20, or is there something more?


EDIT: Just as I've posted I realised that my assumption does not make sense since it is said that exactly one student passes the exam. So I have no clue how to proceed.

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    $\begingroup$ You don't need the probability of all three passing. You need the probability of exactly one passing. There are three ways that can happen. Compute the probabilities for each of the three ways and sum. $\endgroup$
    – lulu
    Sep 16, 2021 at 22:53
  • $\begingroup$ @lulu 0.66+0.75+0.80 is larger than 1, and shouldn't probability be less or equal than 1? $\endgroup$
    – Puki
    Sep 16, 2021 at 22:57
  • $\begingroup$ The sum of those probabilities has nothing to do with anything in the problem. The sum can be greater than $1$ since the events are not disjoint. $\endgroup$
    – lulu
    Sep 16, 2021 at 23:00
  • $\begingroup$ When exactly one passes, exactly two fail, so you need to account for that too.$$\mathsf P((A_1\cap A_2^\complement\cap A_3^\complement)\cup(A_1^\complement\cap A_2\cap A_3^\complement)\cup(A_1^\complement\cap A_2^\complement\cap A_3))$$ $\endgroup$ Sep 16, 2021 at 23:01

2 Answers 2

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$p_1, p_2, p_3 = (2/3), (3/4), (4/5),$ respectively, for the 1st, 2nd, and 3rd students respectively.

$q_1, q_2, q_3 = 1 - p_1, 1 - p_2, 1 - p_3,$ respectively.

Although this is a conditional probability problem, it is also a problem that admits intuitive shortcuts.

Given the constraints, let
$A_1 = p_1 \times q_2 \times q_3,$
$A_2 = q_1 \times p_2 \times q_3,$
$A_3 = q_1 \times q_2 \times p_3.$

Let $A = A_1 + A_2 + A_3.$

Then, the chance of student $k$ being the one that passed is $\displaystyle \frac{A_k}{A} ~: ~k \in \{1,2,3\}.$

$\displaystyle A_1 = (2/3) \times (1/4) \times (1/5) = (2/60).$
$\displaystyle A_2 = (1/3) \times (3/4) \times (1/5) = (3/60).$
$\displaystyle A_3 = (1/3) \times (1/4) \times (4/5) = (4/60).$

Then $\displaystyle A = (9/60).$

$\displaystyle \frac{A_1}{A} = (2/9).$

$\displaystyle \frac{A_2}{A} = (3/9).$

$\displaystyle \frac{A_3}{A} = (4/9).$

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  • $\begingroup$ It absolutely makes sense when described like this, yet I wasn't able to make a connection between some students passing or failing, and then tying those events together. Cheers! $\endgroup$
    – Puki
    Sep 16, 2021 at 23:23
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You can only add the probabilities of events that are mutually exclusive, meaning that they can't both happen at the same time.

The event in which exactly one student passes is made up of three mutually exclusive events:

  1. The first student passes and the other two fail.
  2. The second student passes and the other two fail.
  3. The third student passes and the other two fail.

Notice the importance of adding "and the other two fail" to those events, because the events "the first student passes" and "the second student passes" are not mutually exclusive, they overlap in the event "the first two students pass".

Now, because each student's success is independent of the others, you can figure out the probabilities of those three events by multiplying the individual probabilities - for example, event 1 (where only the first student passes) is the intersection of the events "student 1 passes", "student 2 fails" and "student 3 fails", so its probability is $P(A1 \land A2^c \land A3^c) = P(A1) P(A2^c) P(A3^c) = (2/3)(1/4)(1/5) = 1/30$.

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