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# S1 question that's bugging me watch

1. Right, I have S1 tomorrow and I've been doing a past paper...yet I can't quite understand this one, I have the answer from the mark scheme.

P(A) = 0.65 P(AUB)=0.93

Evaluate P(B) given that:
a) A & B are independent.

So I'd do P(AUB)=P(A)+P(B)-P(A)P(B)
then put in the figures...
P(AUB)=0.65+P(B)-0.65(PB)

And this is where I get confused...
The rest of the mark scheme has:
0.35P(B) = 0.28
P(B) = 0.28

I don't understand where the 0.35P(B) has come from!
Any help is appreciated, thanks!
2. Anyone?
3. 0.65 + 0.28 = 0.93

The question has used P(AUB) = P(A) + P(B)
4. sorry i don't actually think that what you were asking.. i'll have another look
5. sorry i don't actually think that what you were asking.. i'll have another look
6. (Original post by danielharris627)

So I'd do P(AUB)=P(A)+P(B)-P(A)P(B)
then put in the figures...
P(AUB)=0.65+P(B)-0.65(PB)

And this is where I get confused...
The rest of the mark scheme has:
0.35P(B) = 0.28
P(B) = 0.28

I don't understand where the 0.35P(B) has come from!
Any help is appreciated, thanks!
You have P(B)-0.65P(B) in your solution. 1-0.65=0.35 hence the 0.35P(B)

I assume the last line should be P(B) = 0.28/0.35 = 0.8
7. yeah, i agree,

i get how they find 0.35B=0.28
but not sure how they then deduce that B=0.35

sorry :/
8. (Original post by tiny hobbit)
You have P(B)-0.65P(B) in your solution. 1-0.65=0.35 hence the 0.35P(B)

I assume the last line should be P(B) = 0.28/0.35 = 0.8

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