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Sometimes it's quite confusing when it comes to determining how to answer a probability question. Confusions always arise as of whether I should multiply/add or make conditional the probabilities. For example the following:

Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 15% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 10% while that the mother contracted the disease is 9%. What is the probability that both contracted influenza expressed as a whole number percentage?

Let P(F) = Probability that father catches it; P(M) for mother.

I thought the P(both catch it) = P(F)P(M), but the answer is P(at least 1 catch it)= P(F)+P(M)-P(F AND M) and solve for P(F AND M).

My first question is that: I find it particularly difficult to differentiate between addition or multiplication rule when it comes to probabilities from independent events.

My second question is that: I'm also thinking if I'm to use P(at least 1 catch it)= P(F)+P(M)-P(F AND M), I would have make something like: P(at least 1 catch it)= P(F)P(NOT M)+P(M)P(NOT F)+P(F AND M). But it seems the P(F AND M) from two cases are not equivalent? Aren't these 2 expressions representing the same thing?

My third question, even when I calculate P(at least 1 catch it) = 1-P(both not catching it) = 1-P(NOT F)*P(NOT M), P(at least 1 catch it) does not equal to .15 given in the question. What's wrong with my calculation?

Are there any rules in governing which approach to use when solving a probability problem?

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2 Answers 2

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Let's follow up on GlenB's advice and make those Venn diagrams. We do this below with the heterosexual stereotype colours representing mother sick with red/pink and dad sick with blue.

Venn diagram intro

With the two variables mother and father you can create 4 different disjoint situations.

    father sick  and    not mom sick
not father sick  and        mom sick
    father sick  and        mom sick
not father sick  and    not mom sick

It is with those 4 situations that you can perform additive computations.

Intuitively you want to figure out how much the two situations mom sick and father sick overlap (those two may not need to be jisjoint)

different overlap

Your formula

but the answer is P(at least 1 catch it)= P(F)+P(M)-P(F AND M) and solve for P(F AND M)

Stems from the following algebra

algebra solution

You can compare it to a situation with 4 unknowns (the area's/probability of the 4 disjoint pieces) and you try to figure out the values by means of 4 equations. You know

mom sick
0.09 = P(mom sick & not dad sick) + P(mom sick & dad sick)  

dad sick             
0.10 = P(mom sick & dad sick) + P(not mom sick & dad sick)    

one or more sick           
0.15 = P(mom sick & dad sick) + P(not mom sick & dad sick) + P(mom sick & dad sick)   

total probability must be one   
1.00 = P(mom sick & dad sick) + P(not mom sick & dad sick) + P(mom sick & dad sick) + P(not mom sick & not dad sick)  

One final figure to explain the product and sum rule:

explanation sum and product rule

  • When events are disjoint then you can use summation $$P(A \text{ or } B) = P(A) + P(B)$$ note that 'father sick' and 'mom sick' do not meed to be disjoint events. You still get a sum of those events in your solution, but that is due to algebra where we combine multiple equations.

  • When events are independent then you can use the product $$P(A \text{ and } B) = P(A) \cdot P(B)$$ The independence means that the ratio's of the area's/probabilities are unaffected by the other variable. In the image you see the ratio's of 'mom sick' for different states of 'dad sick' whether or not dad is sick the ratio remains the same.

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    $\begingroup$ I particularly appreciate that you drew them by hand. I think that your answer is even more helpful that way (illustrating that diagrams are very much a simple but important part of thinking about a problem, requiring only very basic tools). $\endgroup$
    – Glen_b
    Oct 27, 2022 at 8:23
  • $\begingroup$ @Glen_b I would have wanted to use a computer, but making drawings on a computer can be nasty. $\endgroup$ Oct 27, 2022 at 8:35
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    $\begingroup$ I think the answer is better for it. It's something I should try to keep in mind. $\endgroup$
    – Glen_b
    Oct 27, 2022 at 8:49
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My first question is that: I find it particularly difficult to differentiate between addition or multiplication rule when it comes to probabilities from independent events.

That's not a question (you don't ask anything), but the answer to what I assume is your implied question is simple: there isn't an addition rule for independent events.

The "addition rule" $P(A \text{ or } B) = P(A)+P(B)$ is for mutually exclusive events.

Draw a Venn diagram, from which it's obvious why there's another term there for non-mutually exclusive events (representing the overlap which gets counted twice, once in A and once in B, whereupon you must then subtract one of the overlaps back off again).

My third question, even when I calculate P(at least 1 catch it) = 1-P(both not catching it) = 1-P(NOT F)*P(NOT M), P(at least 1 catch it) does not equal to .15 given in the question. What's wrong with my calculation?

Note that the multiplication rule requires independence.

Did you make sure the events whose probability you multiplied were independent?


Rules for union ("OR") and intersection ("AND") are:

(i) $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

(ii) $P(A \text{ and } B) = P(A)\times P(B|A)$ $\:$ (General product rule)

If you have mutually exclusive events, the third term on the RHS in (i) is $0$, whence "addition rule for mutually exclusive events".

If you have independent events, the second term on the RHS in (ii) is equal to $P(B)$, whence "multiplication rule for independent events".

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    $\begingroup$ Oh right, thanks for clearing the core basic concepts for me. Further, is P(A or B) defined to include P(A AND B) as well? $\endgroup$
    – Student
    Oct 27, 2022 at 6:39
  • $\begingroup$ The definition of union ("OR") is the usual set-union inclusive-or (events are sets). Did you draw a Venn diagram as suggested in my answer? You should be able to see that the interior of all the region included in either A or B being the case also includes the part covered when both are the case, since that's part of each of them. $\endgroup$
    – Glen_b
    Oct 27, 2022 at 7:32

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