Finding conditional probability of a specific problem.

Question

I have this problem which I can't figure out how to solve. I was wondering if my thought process is right or wrong. Some tips would be helpful!

Jen will call Cathy on Saturday with a 60% probability. She will call Cathy on Sunday with an 80% probability. The probability that she will call on neither of the two days is 10%. What is the probability that she will call on Sunday if she calls on Saturday?

This is a problem from Fundamentals of Probability: A First Course By Anirban DasGupta (page 39 problem 3.2)

So I think we can use the hierarchical multiplicative formula.

$P(A_1 \cap A_2 \cap \ ...\ \cap A_k) = P(A_1) P(A_2\mid A_1) P(A_3 \mid A_1 \cap A_2) \times ... \times P(A_k \mid A_1 \cap A_2 \cap A_{k-1})$>

For k = 3 and $A_1$ = {the probability that she will ring (which is 0.9)} $A_2$ = the probability that she will ring on Saturday and $A_3$ = the probability that she will call on Sunday we will have $ 0.8 \times 0.6 =0.9 \times 0.6 \times x $ where $x = P(A_3 \mid A_1 \cap A_2) = \frac{8}{9}$ I know my deductions are wrong, but I can't figure out which parts and how to solve this the right way.

The answer is 5/6, according to my knowledge (source: here).

Answer 1

Call $A$ the event that she calls on saturday and $B$ the event that she calls on sunday.

Then the problem is asking for $P(B | A)$, which by definition is: $$P(B | A) = \frac{P(B \cap A) }{ P(A)}.$$

The problem also gives you the values of

• $P(A)$;
• $P(B)$;
• $P(\bar A \cap \bar B)$.

Then you can use:

• the value of $P(\bar A \cap \bar B)$ to deduce $P(A \cup B)$;
• the values of $P(A)$, $P(B)$ and $P(A \cup B)$ to deduce $P(A \cap B)$.

I confirm that the final answer is $P(B | A) = 5/6$.