How do I find this conditional probability?

Question

Peter wants to send a letter to Mary. The likelihood that Peter will write the letter is $80\%$. Given that Peter has written the letter, there is $90\%$ of chances that the post office won't lose it. Given that the mail hasn't lost it, the mailman has $95\%$ of chances to delivery it correctly to Mary. Once Mary hasn't received the letter, what is the probability that Peter hasn't written it?

MY ATTEMPT

I have obtained the following relations $\mathbb{P}(P) = 0.8$, $\mathbb{P}(O|P) = 0.9$, $\mathbb{P}(D|O) = 0.95$, where the capital letters correspond to the events in the order that they appear. However I am unable to describe the final event and its corresponding probability. Can anyone help me?

Answer 1

Take it one step at a time. First, we determine what the probability that Mary does not receive a letter, irrespective of the reason why she did not receive the letter. This is simple:

• there's a $20\%$ chance that Peter never sent it in the first place;
• there's a $80\% \cdot 10\% = 8\%$ chance that Peter sent the letter, but the post office lost the letter;
• and finally, there's a $80\% \cdot 90\% \cdot 5\% = 3.6\%$ chance that Peter sent the letter, the post office did not lose the letter, but the mailman did not deliver the letter to Mary.

Adding these up gives a total of $20\% + 8\% + 3.6\% = 31.6\%$ chance that Mary does not receive a letter.

Now we compute the probability that Peter did not send the letter (and hence, Mary does not receive one). This is simple - it was the first event in the above list and occurs with probability $20\%.$

So the desired (conditional) probability is simply $\frac{20\%}{31.6\%} = \boxed{63.3\%}.$

Answer 2

$$P(\lnot P \mid \lnot M) = \frac{P(\lnot P \land \lnot M)}{P(\lnot M)} = \frac{P(\lnot P)}{P(\lnot M)}$$ The numerator is easy to compute from the given information. The denominator can be broken up into cases:

• $\lnot P$
• $P \land \lnot O$
• $P \land O \land \lnot D$

Answer 3

Divide the problem into three cases:

• Peter did not write the letter.
• Peter wrote the letter and it arrived.
• No letter arrived.

The first two can easily be computed, and the second and the last cover all situations so they sum to $1$. Divide the first by the the last: $$ \frac{0.2}{1-0.8\cdot0.9\cdot0.95}\approx 0.6329 $$

Answer 4

I like to show this as a tree diagram. Once the whole tree is drawn, you prune off the parts that don't apply because of the conditions that are set.

There are three levels to the tree: writing, losing, delivering. The writing level has two branches, writing and not writing. You assign the probabilities to each branch: .8 on writing and .2 on not writing.

The writing branch has two branches off of it: not losing(.9) and losing(.1). The not writing branch terminates as is.

The not losing branch has two branches: delivering(.95) and not delivering(.05). The losing branch terminates.

The probability of any outcome (the end of a path) is the product of the probabilities as you travel down the branches.

The condition that Mary did not receive the letter means you prune off the deliver branch.

The numerator of your probability is the outcome that Peter didn't write, so you follow that path to get .2

The denominator is the sum of all possible ways that Mary didn't get a letter: (not write) + (write but lose) + (write and not lose but not delivered) = (.2) + (.8 x .1) + (.8 x .9 x .05)

That is it! Draw the diagram to help you see it.