1. ## Multiple event probability

Hi, for some reason I'm having issues with something I usually find easy. The question is as follows:

events A and B are such that P(A) = 2/5, P(B)= 1/2 and P(A|B') = 4/5

Find

a.) P(A n B')
b.) P(A n B)
c.) P(A u B)
d.) P(A|B)

next, state with reason, whether or not A and B are

a) Mutually exclusive
b) independent

What I've done:

When I try and draw out a Venn diagram (As i find them helpful for displaying data) can get P(A) to equal 4/5 when given B' however when I try and work out P(A n B) for the middle of the Venn diagram it comes out as 0.2, when subtracting that from the probabilities of A and B I end up with 0.2 for A. Which no matter what I do I can't make P(A|B') equal 4/5. So I can make my Venn diagram work and I have no idea whether to use the values without subtracting (A n B) or not. I was wondering if they were independent and there were no cases where they are both chosen so P(A n B) would = 0?

BigGScotty278 (AS level Maths & Further Maths)

2. ## Re: Multiple event probability

Originally Posted by BigGScotty278
Hi, for some reason I'm having issues with something I usually find easy. The question is as follows:

events A and B are such that P(A) = 2/5, P(B)= 1/2 and P(A|B') = 4/5

Find

a.) P(A n B')
b.) P(A n B)
c.) P(A u B)
d.) P(A|B)

next, state with reason, whether or not A and B are

a) Mutually exclusive
b) independent

What I've done:

When I try and draw out a Venn diagram (As i find them helpful for displaying data) can get P(A) to equal 4/5 when given B' however when I try and work out P(A n B) for the middle of the Venn diagram it comes out as 0.2, when subtracting that from the probabilities of A and B I end up with 0.2 for A. Which no matter what I do I can't make P(A|B') equal 4/5. So I can make my Venn diagram work and I have no idea whether to use the values without subtracting (A n B) or not. I was wondering if they were independent and there were no cases where they are both chosen so P(A n B) would = 0?

BigGScotty278 (AS level Maths & Further Maths)

$\displaystyle P(A)=P(A|B)P(B)+P(A|B')P(B')$

$\displaystyle \frac{2}{5}=P(A|B) \frac{1}{2}+\frac{4}{5} \left(1-P(B)\right)=P(A|B)\frac{1}{2}+\frac{4}{5} \cdot \frac{1}{2}$

$\displaystyle \frac{2}{5}=P(A|B)\frac{1}{2}+\frac{2}{5}$

$\displaystyle \frac{1}{2}P(A|B)=\frac{2}{5}-\frac{2}{5}=0$ so

$\displaystyle P(A|B)=0$

see if that lets you make headway.

3. ## Re: Multiple event probability

Hello, BigGScotty278!

$\displaystyle \text{Given: }\:P(A)\,=\,0.4,\;P(B)\,=\,0.5,\;P(A|B')\,=\,0.8$

$\displaystyle \text{Find: }\;(a)\;P(A \cap B') \qquad (b)\;P(A \cap B) \qquad (c)\;P(A \cup B) \qquad (d)\;P(A|B)$

$\displaystyle \begin{array}{ccccc}\text{We have:} & P(A) \:=\:0.4 & P(A') \:=\:0.6 \\ & P(B) \:=\:0.5 & P(B') \:=\:0.5 \end{array}$

$\displaystyle P(A|B') \:=\:0.8 \quad\Rightarrow\quad \frac{P(A\cap B')}{P(B')} \:=\:0.8 \quad\Rightarrow\quad P(A\cap B') \:=\:(0.8)\!\cdot\!P(B')$

. . $\displaystyle P(A \cap B') \:=\:(0.8)(0.5) \quad\Rightarrow\quad P(A\cap B') \:=\:0.4$

With the above information, we have this chart:

. . $\displaystyle \begin{array}{c||c|c||c|} & B & B' & \text{Total} \\ \hline\hline A && 0.4 & 0.4 \\ \hline A' & && 0.6 \\ \hline\hline \text{Total} & 0.5 & 0.5 & 1.0 \\ \hline \end{array}$

We can complete the chart:

. . $\displaystyle \begin{array}{c||c|c||c|} & B & B' & \text{Total} \\ \hline\hline A &0.0& 0.4 & 0.4 \\ \hline A' & 0.5 & 0.1 & 0.6 \\ \hline\hline \text{Total} & 0.5 & 0.5 & 1.0 \\ \hline \end{array}$

. . $\displaystyle (a)\;P(A\cap B') \:=\:0.4$

. . $\displaystyle (b)\;P(A \cap B) \:=\:0$

. . $\displaystyle (c)\;P(A\cup B) \:=\:0.9$

. . $\displaystyle (d)\;P(A|B) \:=\:\frac{P(A\cap B)}{P(B)} \:=\:\frac{0}{0.5} \:=\:0$

$\displaystyle \text{Next, state with reason, whether or not }A\text{ and }B\text{ are:}$

. . $\displaystyle \text{(a) Mutually exclusive} \qquad \text{(b) independent}$

$\displaystyle \text{(a) Since }P(A\cap B) = 0,\:A\text{ and }B\:are\text{ mutually exclusive.}$

$\displaystyle (b) \;P(A)\!\cdot\!P(B) \:=\:(0.4)(0.5) \:=\:0.2$
. .$\displaystyle P(A \cap B) \:=\:0$

$\displaystyle \text{Since }P(A)\!\cdot\!P(B) \,\ne\,P(A\cap B),\,A\text{ and }B\text{ are }not\text{ independent.}$