# Basic probability.. has both a1 and a2 but not a3.

A system can have three different types of defects: A1, A2, A3. We are given the following probabilities.

P(A1) = .12   P(A2) = .07    P(A3) = .05
P(A1 U A2) = .13   P(A1 U A3) = .14
P(A2 U A3) = .10   P(A1 and A2 and A3) = .01


What is the probability that the system has both type 1 and type 2 defects, but not a type 3 defect.

I'm confusing the heck out of myself with this one. I've drawn out a Venn Diagram to try and help visualize what we want, A1 and A2 are the left and right circles, A3 is the bottom circle. We are looking for A1, A2, and everything not A3 to be shaded. So we need to do the following:

P(A1) + P(A2) + P(A3') - P(A1 U A2) - P(A1 U A3)- P(A2 and A3)  = + P(A1 and A2 and A3)

.12 + .07 + .05 - .13 - .14 - .10 + .01 = -.12


-.12 is clearly not right. Can someone please shed some light on this for me?

Update: Can we do something like 1 - P(A3)? This would give us everything excluding A3.

If you are going to down vote the question, at least provide some feedback so I can improve the question.

We are tasked with finding $Pr(A_1\cap A_2\cap A_3^c)$, the event where specifically the first two defects occur while the third defect does not occur.

sidenote: here I use $A_3^c$ to refer to the complementary event to $A_3$

We know that $Pr(A_1\cap A_2\cap A_3^c)+Pr(A_1\cap A_2\cap A_3) = Pr(A_1\cap A_2)$

This is because $A_1\cap A_2\cap A_3^c$ and $A_1\cap A_2\cap A_3$ are disjoint events whose union is $A_1\cap A_2$. (this follows immediately from inclusion-exclusion which is stated below)

We are told $Pr(A_1\cap A_2\cap A_3)$. The only missing piece of information is $Pr(A_1\cap A_2)$

Inclusion-Exclusion: (two event case) $$Pr(A)+Pr(B)=Pr(A\cup B)+Pr(A\cap B)$$

We are told $Pr(A_1), Pr(A_2)$, and $Pr(A_1\cup A_2)$. Use this to find $Pr(A_1\cap A_2)$ and plug into what we found above.

$Pr(A_1\cap A_2) = Pr(A_1)+Pr(A_2)-Pr(A_1\cup A_2) = 0.12+0.07-0.13 = 0.06$

.

$Pr(A_1\cap A_2\cap A_3^c)=Pr(A_1\cap A_2)-Pr(A_1\cap A_2\cap A_3) = 0.06 - 0.01 = 0.05$

As a side note, it appears that you tried to use inclusion exclusion in your work above incorrectly. The three-event inclusion-exclusion should read as:

$$Pr(A\cup B\cup C)=Pr(A)+Pr(B)+Pr(C)-Pr(A\cap B)-Pr(A\cap C)-Pr(B\cap C)+Pr(A\cap B\cap C)$$

However the following is false:

$$\color{red}{Pr(A\cap B\cap C)=Pr(A)+Pr(B)+Pr(C)-Pr(A\cup B)-Pr(A\cup C)-Pr(B\cup C)+Pr(A\cup B\cup C)}$$

It would be for this reason that you got a nonsensical probability in your earlier attempt. (Unless working in a highly unusual setting, negative probabilities are impossible)

• Thanks for the help. Is there any way possible that you could draw out a venn diagram to show this. I'm not really understanding how this gives us our answer. Why is it not simply A1 * A2 * A3'? $Pr(A_1\cap A_2\cap A_3^c)=Pr(A_1\cap A_2)-Pr(A_1\cap A_2\cap A_3) = 0.06 - 0.01 = 0.05$ – Talen Kylon Dec 15 '15 at 6:29
• @TalenKylon That would be highly tedious, so no. Here is wolfram's picture of the venn diagram, but without the values entered in. You can find all values by applying inclusion-exclusion on your own. – JMoravitz Dec 15 '15 at 6:33
• As for your question why is it not simply A1 * A2 * A3', what do the *'s represent? – JMoravitz Dec 15 '15 at 6:35
• multiplication for and – Talen Kylon Dec 15 '15 at 6:36
• $Pr(A_1)\cdot Pr(A_2) = Pr(A_1\cap A_2)$ if and only if $A_1$ and $A_2$ are Independent Events. That is not the case here. – JMoravitz Dec 15 '15 at 6:38