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Given that $P(A)=0.4$ and $P(B|A)=0.5$. Hence find find $P(A' \cup B')$. How to do this one?

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    Why is it not 0.5...? (If they are disjoint) – 2017-01-01
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    Don't know nothing more given in that problem – 2017-01-01
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    What does "**After A has occurred** find $\rm P(A'\cup B')$" mean? – 2017-01-01
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    Question edited that is actually probability of B after A has occurred is 0.5 – 2017-01-01
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    Shouldn't you then write $P(B \vert A) = 0.5$ then? – 2017-01-01
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    Yeah I know that can you give solution to problem please – 2017-01-01
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    If you know that, then edit and write that. Does P (x $\cup $ y) mean the probability of either one or the other, or the probability of both? – 2017-01-01
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    See if my edit change your question then revert my changes. – 2017-01-01
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    @KanwaljitSingh No its right thanks – 2017-01-01
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    Your welcome... – 2017-01-01
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    What happened? Was there independent? – 2017-01-01
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    @Bellmondo Though your answer is also right but there is nothing given about their independences in the book , but assuming them independent worked in my case, thanks for the help. – 2017-01-01
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    @ShubhamWagh Your Welcome... There is something wrong in your question. If it was $P(A|B)$ You need not independency. But if it is $P(B|A)$ you need it. – 2017-01-01
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    @Bellmondo I guess there is misprint in book . thanks I will remember it . – 2017-01-01

2 Answers 2

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$P(A|B)=\frac{P(A , B)}{P(A)}$

So $P(A,B)=0.2$ and the answer is $0.8$

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Use formula-

$P(A'\cup B') = 1- P(A\cap B)$

Here's how it drive.

$P(A'\cup B') = P(A\cap B)'$

= $1- P(A\cap B)$

Also P(B|A) = 0.5

$\frac{P(A\cap B)}{P(A)}$ = 0.5

So $P(A\cap B)$ = 0.5 * 0.4 = 0.20

So in your case answer,

$P(A'\cup B') = 1- P(A\cap B)$

= 1 - 0.20 = 0.80

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    Are you sure that P(A intersection B) is equal to P(A)P(B)? – 2017-01-01
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    Also it is P(B/A) =0.5 – 2017-01-01
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    If these events are independent then yes. – 2017-01-01
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    Ok I read it wrong. Then edit your questions. – 2017-01-01