Danyal124
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Part c ver confused I thought if u restrict it to p(A’)=0.5 and P(B)=0.2 in a’ so 0.2/0.5 is still wrong ??? Any help?Name:  99F31E2E-E87A-4265-B54F-68D4AF0D89E6.jpeg
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old_engineer
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(Original post by Danyal124)
Part c ver confused I thought if u restrict it to p(A’)=0.5 and P(B)=0.2 in a’ so 0.2/0.5 is still wrong ??? Any help?
This is conditional probability. P(B given A') = P(B n A') / P(A'). Standard formula that you should know if you're tackling this kind of question.
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S0FT
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a) P(a) = 1 - 0.3 - 0.2 - 0.1 = 0.4
b) P(b) = 0.3 + 0.1 = 0.4
c) just use a conditional probability and apply the formula :
P(Win2 given Win1’) = P(Win2 intersect Win1’) / P(Win1’) = 0.1 / 0.5 = 0.2.
d) if the events are independent then, P(Win2 intersect Win1) = P(Win1) * P(Win2).
P(Win2 intersect Win1) = 0.2 and P(Win1) * P(Win2) = 0.15
As a result we can conclude the events are not independent.

I hope I helped
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old_engineer
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(Original post by S0FT)
a) P(a) = ....
Please observe the posting guidelines (sticky): hints not full solutions. Please delete your full solution.
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davros
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(Original post by S0FT)
a) P(a) =

I hope I helped
Please don't post full solutions - it's against the rules of the forum
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Danyal124
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(Original post by S0FT)
a) P(a) = 1 - 0.3 - 0.2 - 0.1 = 0.4
b) P(b) = 0.3 + 0.1 = 0.4
c) just use a conditional probability and apply the formula :
P(Win2 given Win1’) = P(Win2 intersect Win1’) / P(Win1’) = 0.1 / 0.5 = 0.2.
d) if the events are independent then, P(Win2 intersect Win1) = P(Win1) * P(Win2).
P(Win2 intersect Win1) = 0.2 and P(Win1) * P(Win2) = 0.15
As a result we can conclude the events are not indepei dont ndent.

I hope I helped
i dont understand how to get P(BnA') for part c could you explain it please
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old_engineer
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(Original post by Danyal124)
i dont understand how to get P(BnA') for part c could you explain it please
I suggest that you draw a Venn diagram and mark it up with what you know. Post your working (including diagram) if still stuck.
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