Question:
Parcels from sender 'S' to receiver 'R' pass sequentially through 2 post-offices.
Each post-office has probability 1/5 of losing an incoming parcel, independently of all other parcels.
Given that a parcel is lost, what is the probability that it was lost by the second post-office.
Why do we have to use conditional probability in this problem?
Shouldn't the answer just be:
P(lost at 2nd) = P(not lost at 1st) * P(lost at 2nd)
P(lost at 2nd) = 4/5 * 1/5
P(lost at 2nd) = 4/25
But the answer is 4/9
Given your question, see if this helps.
Say there are $100$ letters sent.
You expect first post office to lose $20$ letters $( = \frac{1}{5} \times 100)$.
You expect second post office to lose another $16$ letters $( = \frac{1}{5} \times (100-20)$).
So we expect $64$ letters to reach and $36$ letters to be lost.
Now we are told that our sample space is no more $100$ letters. It is those $36$ letters that are lost.
Probability that it is Post office $2$ that loses the letter if the letter is lost
$\displaystyle = \frac{16}{36} = \frac{4}{9}$