# Math Help - Question on Probability

1. ## Question on Probability

Given that
$P(\neg A)=0.6$
$P(B\mid A)=0.7$
$P(B)=0.3$
What is $P(\neg A \wedge B)$?

2. Originally Posted by quiney
Given that
$P(\neg A)=0.6$
$P(B\mid A)=0.7$
$P(B)=0.3$
What is $P(\neg A \wedge B)$?
A simple approach would be to draw a tree diagram. Have you tried doing that?

3. Here is another way.
Recall that $P(B) = P(B \cap A) + P(B \cap \neg A)$
Solve for $P(B \cap \neg A)$.

4. Hello, quiney!

Yet another approach . . .

Given that:
. . $\begin{array}{ccc}P(\sim\!A)&=& 0.6 \\
P(B\,|\,A) &=& 0.7 \\
P(B) &=& 0.3 \end{array}$

What is . $P(\sim\!A \wedge B)$ ?

We can place the data into a chart:

. . $\begin{array}{c||c|c||c}
& B & \sim\!B & \text{total} \\ \hline \hline
A & & & 0.40 \\ \hline
\sim\!A & & & 0.60 \\ \hline \hline
\text{Total} & 0.30 & 0.70 & 1.00 \end{array}$

We have: . $P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(B \wedge A)}{P(A)} \:=\:0.7$

. . . . $P(B \wedge A) \:=\:0.7\cdot P(A) \;=\;(0.7)(0.4)$

. . . . $P(A \wedge B) \;=\;0.28$

Insert that into the chart.
. . Fill in the rest of the chart.

. . $\begin{array}{c||c|c||c}
& B & \sim\!B & \text{total} \\ \hline \hline
A & 0.28 & 0.12 & 0.40 \\ \hline
\sim\!A & 0.02 & 0.58 & 0.60 \\ \hline \hline
\text{Total} & 0.30 & 0.70 & 1.00 \end{array}$

Therefore: . $P(\sim\!A \wedge B) \;=\;0.02$

Corrected my typo . . . Thanks, Plato!

5. Originally Posted by Soroban
We have: . $P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(A \wedge B)}{P(B)} \:=\:0.7$
Please note the typo $\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}$.

6. Thanks a lot everybody. So, since $p(B)=p(B \wedge A) + p(B \wedge \neg A)$:
$0.3=p(B \wedge A) + p(B \wedge \neg A)$

Since $p(B|A)=0.7=\dfrac{p(B \wedge A)}{p(A)=0.4} \quad\Rightarrow\quad\ 0.28=p(B\wedge A) \quad\Rightarrow\ 0.3=0.28 + p(B \wedge\neg A)$
Therefore
$0.02=p(B \wedge\neg A)=p(\neg A \wedge B)$

Funny, the answer guide in the book says 0.2. Must be a typo.