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Mathematics 8 Online
OpenStudy (anonymous):

If A and B are dependent events, which of these conditions must be true? P(A and B) = P(A) + P(B) P(A and B) = P(B) P(A) P(B|A) = P(B) P(A|B) = P(A) P(B|A) P

OpenStudy (anonymous):

@YanaSidlinskiy

OpenStudy (yanasidlinskiy):

Have you tried it on your own?

OpenStudy (anonymous):

I feel that it is A but I am not sure :/

OpenStudy (yanasidlinskiy):

Sorry for long replies...I'm having technical issues right now. But do you see where I'm going with this, right? If A and B are dependent events and A happens first you'd get something like this: P(A, B) = P(A) • P(B when A has happened) Which will give you: P(A , B) = P(A) • P(B|A)

OpenStudy (perl):

is there supposed to be a fraction bar on choice b)

OpenStudy (anonymous):

I'm kind of lost. The independent and dependent questions throw me off.

OpenStudy (anonymous):

Yes perl a fraction bar on all choices

OpenStudy (anonymous):

is it B?

OpenStudy (yanasidlinskiy):

I kind of gave you an answer....

OpenStudy (anonymous):

P ( A and B ) = p(A) * P(B)?

OpenStudy (anonymous):

Perl, What do you think?

OpenStudy (yanasidlinskiy):

Sorry if it's really confusing you. I'm really not sure how to explain it to you....I'm not great at explaining but here's like something you should do to eliminate your options down: P(A|B) = y A is y WRONG P(B|A) = y Probability of B, (when A has happened), is y "the probability of event B is y"

OpenStudy (anonymous):

C?

OpenStudy (perl):

can you post the question with fraction bars?

OpenStudy (perl):

I want to be sure I am reading this correctly .

OpenStudy (perl):

this is what i see If A and B are dependent events, which of these conditions must be true? P(A and B) = P(A) + P(B) P(A and B) = P(B) P(A) P(B|A) = P(B) P(A|B) = P(A) P(B|A) P

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