Intersections

Let A and B be events in a sample space S such that P(A) = 0.6, P(B) = 0.5, and P(A intersection B) = 0.25. Find the probabilities below. Hint: (A intersection Bc) union (A intersection B) = A
(a) P(A|B^c)=.7

(b) P(B|A^c)

Can you help me with b?

Hello, sportywarbz!

Let \(\displaystyle A\) and \(\displaystyle B\) be events in a sample space \(\displaystyle S\) such that: .\(\displaystyle P(A) = 0.6,\;\; P(B) = 0.5,\;\;P(A \cap B) = 0.25.\)

Click to expand...

Insert the given information into a chart:

. . \(\displaystyle \begin{array}{c||c|c||c|} & P(B) & P(B^c) & \text{Total} \\ \hline \hline P(A) & 0.25 && 0.60 \\ \hline P(A^c) &&& \\ \hline\hline \text{Total} & 0.50 &&\\ \hline \end{array}\)


Complete the chart:

. . \(\displaystyle \begin{array}{c||c|c||c|} & P(B) & P(B^c) & \text{Total} \\ \hline \hline P(A) & 0.25 & 0.35 & 0.60 \\ \hline P(A^c) & 0.25 & 0.15 & 0.40 \\ \hline\hline \text{Total} & 0.50 & 0.50 & 1.00 \\ \hline \end{array}\)


Then apply Bayes' Theorem.

Find the following probabilities:

. . \(\displaystyle (a)\;\;P(A|B^c)\)

Click to expand...

\(\displaystyle P(A\,|\,B^c) \;=\;\frac{P(A \cap B^c)}{P(B^c)} \;=\;0.35}{0.50} \:=\:0.7\)

Your answer is correct!

\(\displaystyle (b)\;\;P(B|A^c)\)

Click to expand...

\(\displaystyle P(B\,|\,A^c) \;=\;\frac{P(B \cap A^c)}{P(A^c)} \;=\;\frac{0.25}{0.40} \;=\;0.625\)