Using Bayes' Theorem with three events

Question

As a short seatwork in probability, we were given this problem but I am not exactly sure if we're supposed to use Bayes' Theorem for this one:

If you get to have a nap for at least 3 hours, then you can stay awake at midnight 80% of the time. If you get 1-3 hours of nap, then you can stay awake at midnight 60% of the time. Lastly, if you get 1 hour or less, then you can only stay awake at midnight 20% of the time.

Let there be 10% chance that you get to nap for at least 3 hours, 70% chance to nap 1-3 hours, and 20% chance to nap for at most 1 hour.

If you fell asleep through the whole night, what is the probability that you got to nap for at least 3 hours?

How do we solve this kind of problem?

Answer 1

Some hints:

First formalize by defining events:

• $A$ - stays awake $\Rightarrow A'$ sleeps through
• $E_1$ - gets at most 1-hour nap
• $E_2$ - gets 1-3 hours nap
• $E_3$ - gets at least 3-hours nap

Now, the data given is:

• $P(E_1)=0.2$, $P(E_2)=0.7$, $P(E_3)=0.1$
• $P(A|E_1) = 0.2$, $P(A|E_2) = 0.6$, $P(A|E_3) = 0.8$

The question asks for

\begin{eqnarray*}P(E_3|A') & = & \frac{P(E_3\cap A')}{P(A')}\\ & = & \frac{P(E_3) - P(E_3\cap A)}{1-P(A)} \\ & \stackrel{P(E_i \cap A)=P(A|E_i)P(E_i)}{=} & \frac{P(E_3) - P(A|E_3)P(E_3)}{1-\sum_{i=1}^3 P(A|E_i)P(E_i)} \end{eqnarray*}

Can you take it from here?