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.
Best Answer
$p_1, p_2, p_3 = (2/3), (3/4), (4/5),$ 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 $\displaystyle A = (9/60).$
$\displaystyle \frac{A_1}{A} = (2/9).$
$\displaystyle \frac{A_2}{A} = (3/9).$
$\displaystyle \frac{A_3}{A} = (4/9).$