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Mathematics 12 Online
OpenStudy (anonymous):

The probability of Dawn reading before going to bed is 5/6. The probability of Dawn brushing her teeth before going to bed is 6/7. The probability of Dawn watching TV before going to bed is 1/5. What is the probability that she will do all three activities before going to bed? A.2/3 B.5/7 C.2/35 D.1/7

OpenStudy (anonymous):

making the rather broad and erroneous assumption that these events are independent, you can multipy the probabilities together

hero (hero):

Hint: When three events, A and B and C, are independent, the probability of all three occurring is: P(A and B and C) = P(A) · P(B) · P(C)

hero (hero):

Why wouldn't they be independent?

hero (hero):

@satellite73

OpenStudy (anonymous):

30/210.... so that needs to be reduce?

hero (hero):

30/210 = 3/21 = 1/7

OpenStudy (anonymous):

oh.... ok that makes sense. Thanks!

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