Probability problem Bayes

Question

In 80% of the cases a worker uses the subway to get to work. If he gets it there's an 3/4 probability to get to work on time. In average 4 out of 6 days he gets to work on time.Today he didn't get to work. Whats the probability he took the subway?

I was thinking I could have two events A:"He gets to work at time" B:"He takes the subway" and maybe use Bayes but I'm not sure how to.

Answer 1

Let $A = \{\text{arrive on time}\}\, B = \{\text{takes train}\}$. Then you are interested in calculating the event

$$\mathbf{P}[ B \, | \, A^c]$$

where $A^c$ is the complement of $A$,i.e. $A^c = \{\text{does not arrive on time}\}$.

From the information you give above

\begin{align*} \mathbf{P}[B] & = \frac45 \\ \mathbf{P}[A \, | \, B] & = \frac34 \\ \mathbf{P}[A] & = \frac46 = \frac23 \end{align*} and from these we can deduce \begin{align*} \mathbf{P}[A^c \, | \, B] & = 1-\mathbf{P}[A \, | \, B] = \frac14 \\ \mathbf{P}[A^c]\ &= 1- \mathbf{P}[A] =\frac13 \end{align*}

Now we are in a position to use Bayes Rule

\begin{align*} \mathbf{P}[B \, | \, A^c] & = \mathbf{P}[A^c \, | \, B] \frac{\mathbf{P}[B]}{\mathbf{P}[A^c]} \\ & = \left( \frac14 \times \frac45 \right) \big/ \left( \frac13 \right) \\ & = \frac35 \end{align*}

Answer 2

HINT:

Letting A and B be as you described, we have \begin{equation} P(A|B) = 3/4, \end{equation}

and we are interested in \begin{equation} P(B|A^c) = \frac{P(A^c\cap B)}{P(A)}. \end{equation}

Recall that for any two events X, Y, \begin{equation} P(X) = P(X \cap Y) + P(X \cap Y^c). \end{equation}