Why Conditional Probability is used 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
 A: P(lost somewhere) $=1-(4/5)^2=9/25$
P(lost at second | lost somewhere) $=\frac{4/25}{9/25}=4/9$
A: You need conditional probability because of the way the question was phrased.
You also need to actually understand the question. It's not about the probability that a package gets lost, nor the that a randomly chosen package gets lost by the second post office. It's about only those parcels that actually are lost.
$$
\begin{array}{|ccccccc|}
\hline
& & \text{lost by 1st p.o.} & & & & \frac 1 5 = \frac 5 {25} \\ & \nearrow \\ \text{start} & & & & & \text{lost by 2nd p.o.} & \frac 4 5 \times \frac 1 5 = \frac 4 {25} \\
& \searrow & & & \nearrow \\
& & \text{not lost by 1st p.o.} \\
& & & & \searrow \\
& & & & & \text{not lost by 2nd p.o.} & \frac 4 5 \times \frac 4 5 = \frac{16}{25} \\ \hline
\end{array}
$$
The probability of being lost is $\dfrac{5+4}{25}.$
Among those $9$ cases out of $25$ that are lost, $5$ are lost by the first post office and $4$ by the second.
Or you can express it like this: Suppose $25$ parcels are mailed. The $5$ are lost by the first post office, so $20$ go on to the second. Of those $20$ the number lost by the second post office is $4$. Thus $9$ parcels have been lost: $5$ by the first post office and $4$ by the second.
A: 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
$\displaystyle  = \frac{16}{36} = \frac{4}{9}$
