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lets say I have a switch A with 3 legs, each leg has 0.8 chance to be connected (and then electricity will flow), we need only 1 leg connected for A to transfer the electricity (so sorry I didn't explain it that well I'm having hard time to translate this problem)

So I calculated the chance of A to transfer electricity by doing $1-(0.2)^3$ which is $124\over125$ which I think is true.
The problem is I wanted to say that A will transfer elecricity only if 1 of his legs will be connected so its like saying $0.8 + 0.8 + 0.8$ which is obviously wrong(over $1$) so I used the weird formula that says to do like this: $0.8 + 0.8 + 0.8 - (0.8)^2 -(0.8)^2 - (0.8)^2 + (0.8)^3 = {124\over125}$ too.

My only problem is that I didnt understand why I had to use that formula and why I could multiple probabilities for the intersection but couldn't just sum them for the union.

Thanks in advance

The One
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3 Answers3

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The Product Rule applies to events which are independent. Only then is the probability of the intersection equal to the product of the probabilities of the events.

$\mathsf P(A\cap B) ~=~ \mathsf P(A)~\mathsf P(B)~$ only when events $A$ and $B$ are independent.

(When dealing with more than two events we require mutual independence.)

Otherwise conditional probability must be used: $\mathsf P(A\cap B)~=~\mathsf P(A)~\mathsf P(B\mid A)\\\qquad\qquad~=~\mathsf P(A\mid B)~\mathsf P(B)$

The Addition Rule applies only when the events are mutually exclusive (also known as disjoint).   Only then is the probability of the union equal to the sum of probabilities of the event.

$\mathsf P(A\cup B)~=~\mathsf P(A)+\mathsf P(B)$

Otherwise if the events are not disjoint (ie they have common outcomes) then we would be over measuring and must exclude the measure of the intersection.

$\mathsf P(A\cup B)~=~\mathsf P(A)+\mathsf P(B) - \mathsf P(A\cap B)$

When dealing with more than two events, the principle of inclusion and exclusion is required

$\begin{align}\mathsf P(A\cup B\cup C)~=~&\mathsf P(A)+\mathsf P(B)+\mathsf P(C) - \mathsf P(A\cap B)-\mathsf P(A\cap C)-\mathsf P(B\cap C)+\mathsf P(A\cap B\cap C)\end{align}$

... and so on.

$\Box$

Graham Kemp
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This is the inclusion-exclusion principle. We have that $$ \mathbb{P}(A_1\cup A_2\cup A_3)=\mathbb{P}(A_1)+\mathbb{P}(A_2)+\mathbb{P}(A_3)-\mathbb{P}(A_1\cap A_2)-\mathbb{P}(A_1\cap A_3)-\mathbb{P}(A_2\cap A_3)+\mathbb{P}(A_1\cap A_2\cap A_3) $$ for any three events $A_1$, $A_2$ and $A_3$. The following picture illustrates the idea behind this formula: enter image description here

Cm7F7Bb
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  • when do you use this principle and when you just sum it up? (like coin flip is either 0.5 heads or 0.5 tails so if you do a union you indeed get 1) – The One Apr 15 '16 at 11:30
  • @TheOne The probability of the union of events equals the sum their probabilities only when the events are mutually exclusive (also known as disjoint). IE: they can not have any common outcomes - meaning all the probabilities of intersections are zero. – Graham Kemp Apr 15 '16 at 11:34
  • @TheOne If the events are mutually exclusive (i.e. they cannot happen at the same time), then $P(A_1\cup A_2\cup A_3)=P(A_1)+P(A_2)+P(A_3)$. If they are not mutually exclusive, we have to use this principle. – Cm7F7Bb Apr 15 '16 at 11:35
  • and when does we can't say that $P(A \intersection B)=P(A)*P(B)$? I think there is another formula for that too I forgot what it looks like and when do we use it – The One Apr 15 '16 at 11:42
  • @TheOne If two events $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)$. The legs of the switch A are independent in your example. – Cm7F7Bb Apr 15 '16 at 11:43
  • Can you give me an example of dependent events? thanks – The One Apr 15 '16 at 11:44
  • @TheOne Suppose that we roll a fair die. The sample space is then given by $\Omega={1,2,3,4,5,6}$. Consider two events: $A={\text{the outcome is greater than 3}}$ and $B={\text{the outcome is an even number}}$. The events $A$ and $B$ are not independent. – Cm7F7Bb Apr 15 '16 at 11:47
  • @TheOne Toss a die. The events A:"show an even number (ie 2,4,6)" and B:"show a prime number (ie 2,3,5)" are dependent. $P(A)=1/2, P(B)=1/2, P(A\cap B) = 1/6$ – Graham Kemp Apr 15 '16 at 11:48
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Hint One of the three legs running isnt dependent on the running of other two legs. So the events are independent events. So we just multiply their probabilities instead of adding .

Archis Welankar
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