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Suppose a student takes 3 tests on a given day and that the probability that she passes any particular test is .6, the probability that she passes any particular pair of tests is .4, and the probability that she passes all 3 tests is .3. What is the probability that she passes at least one test ?

My idea to this question is to model it in terms of P(A) = probability of passing test A with tests A, B , C. So P(passes at least one test) = 1 - P(passing no tests)

P(passing no tests) = P((AUBUC)') = 1 - P(AUBUC)
P(AUBUC) = inclusion exclusion theory which is = 0.6 * 3 - 0.4 * 3 - 0.3 = 0.3
P(passing no tests) = 1 - 0.3 = 0.7

Obviously I'm missing something here, anyone can help me with this one?

utobi
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sasaasasasha
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    Don't apply formulas, especially when you run the risk of getting a sign wrong: draw the Venn diagram. The seven pieces of information (three individual tests, three pairs of tests, and one triple of tests) along with the probability axioms permit you to determine the probabilities of all 8 component events (corresponding to individual regions in the diagram), thereby answering probability questions about any event. – whuber May 11 '23 at 15:08
  • $P(A\cup B \cup C) \ge P(A)=0.6$ so saying it is $0.3$ must be wrong. Perhaps you have a sign wrong in your inclusion-exclusion? – Henry May 11 '23 at 15:18
  • @Henry Well it is actually a question in a maths book.... The question is as it is like the first paragraph ^. And the answer provided says its .9 to which I'm, not sure. Thanks for getting back to me though. I tried drawing out a venn diagram too, but it doesnt make sense to me that end up my venn diagram shows that the intersections are greater than the P(any test) in the diagram – sasaasasasha May 11 '23 at 15:24
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    Follow @whuber's suggestion of a Venn diagram and you can check $0.9$ is correct – Henry May 11 '23 at 15:26
  • I don't really understand after looking at the venn diagram too. I'm supposed to find the probability of the area outside the venn diagram and that $ 1 - AreaOutside$ – sasaasasasha May 11 '23 at 15:45
  • Emulate the procedure shown at https://stats.stackexchange.com/a/466505/919. In particular, begin with the inner probability (the triple intersection, equal to 0.3 here) and work outwards. – whuber May 11 '23 at 15:53
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    @whuber Omg I did not realise it could be drawn like that. Thank you so much! – sasaasasasha May 11 '23 at 15:56

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