C3_W1 sum of probabilities joint events

in the video the introductory quiz : 0.6 of kids play soccer and 0.5 play basket. How many play soccer or basket? and the answer is : we do not have enough information.
But what about this answer: 0.6+0.5-0.6*.05=0.8 is ratio of kids who play either or both. Further 0.3 play soccer only and 0.2 play basket only and 0.3 play both and 0.2 do not play basket or soccer. We all needed information, right?

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I have not taken this course myself, so I don’t know the context and how deeply they go into probability theory, but the formula you give only works if the two properties (plays soccer and plays basketball) are “independent events”. But that is probably not true, right? Or at least you don’t know that, so it is a true statement that you don’t have enough information to conclude an answer.

It is a good point that the probability of A and B is:

P(A) + P(B) - P(A \cap B)

And if A and B are independent events (which is a technical concept in probability theory), then you have:

P(A \cap B) = P(A) * P(B)

But the point is what I said above: you don’t have enough information to know if A and B are independent in this instance, so you just don’t know the answer here. If they gave you the probability of P(A \cap B), then you would be able to apply your formula.


A search of the forum for the word “soccer” turned up this previous discussion:

Thanks paulinpaloalto. still not clear to me. Instructors solution is the same as solution we can arrive at by assuming sports are independent. Is it legitimate to ask the question what if sports were dependent?