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Intuitive explanation for P(A U B) = P(A) + P(B)

  1. Oct 24, 2015 #1

    I am unsure as to why the probability of A U B = P(A) + P(B) for sets A and B being disjoint. Why do we add the two probabilities? Is it because the sample space remains the same but now we add the numerators as the number of events have increased and that is why we are adding the two?? Can anyone give me a good intuitive explanation? I tried solving the problem myself but I cannot figure out how to arrive at the answer. An intuitive explanation will really help. Thanks in advance for your answers!
  2. jcsd
  3. Oct 24, 2015 #2


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    What reason do you have to doubt this? It's difficult to know how to explain this unless you can describe what you see as the probability of two disjoint events? Let's take an example:

    The probablity a die comes up odd is 1/2. And the probability it comes up 4 is 1/6. So, the probability it is either odd or 4 is 1/2 + 1/6 = 2/3.

    What do you think it should be?
  4. Oct 24, 2015 #3
    Thank you for your answer. I think I understand now. Please do let me know if my explanation is correct. In your example, we are adding the two events (it is odd and number is 4) and grouping them as a single event and hence it is addition. Moreover, the sample space remains the same i.e. 6 alternatives possible. This also brings me to the conclusion that we have to subtract the element counted more than once. Thanks! Do let me know.
  5. Oct 24, 2015 #4


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    As you may know, in general:

    ##P(A \cup B) = P(A) + P(B) - P(A \cap B)##

    Perhaps the simplest way is to think of events as shapes and the probability as the area of the shape and use Venn diagrams.

    In the special case where ##A## and ##B## are disjoint (which means they can't both happen), then ##P(A \cap B) = 0##

    I would tend to think of this in terms of simple examples (like coins, dice and cards) and then abstract that inituitive understanding into the abstract notation of sample spaces and events.
  6. Oct 24, 2015 #5
    @PeroK : Thank you for your explanation. Got me to realize why P(AUB) = P(A) + P(B)!
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