C4 Probability. Another one, Sorry! Watch

Sushmeist
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#1
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Two events R and S are independant. P(R|S) = 3/4. P(S) = P(R'nS').

Find

a) P(S)
b) P(S'nR)
c) Write down P(S|R)

Stuck on a) all ready, keep coming up with 4/7. Answer meant to be 1/5. Can anyone enlighten me???

Help on b) and c) would be great too

Thanks
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CalculusMan
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C4? Are you sure this isn't S1?

I think I remember doing this question (or similar) in S1, I can get my book and post my solution to it if you wish, but I didn't think there was any probability in C4.
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Sushmeist
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You're right S1!!!! Brain seems to have switched off!
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CalculusMan
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If R and S are independent then P(R|S) = P(R) is a hint.

Do you want the full solution?
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Christophicus
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(Original post by Sushmeist)
Two events R and S are independant. P(R|S) = 3/4. P(S) = P(R'nS').

Find

a) P(S)
b) P(S'nR)
c) Write down P(S|R)

Stuck on a) all ready, keep coming up with 4/7. Answer meant to be 1/5. Can anyone enlighten me???

Help on b) and c) would be great too

Thanks
Hi

a) For independent events, P(RnS) = P(R).P(S)
P(R|S) = 3/4
=> P(RnS)/P(S) = 3/4
=> P(R).P(S)/P(S) = 3/4
=> P(R) = 3/4

P(S) = P(R'nS') = 1-P(SuR)

P(SuR) = P(R)+P(S)-P(RnS)

=> P(S) = 1-(P(R)+P(S)-P(RnS))
=> 2P(S) = 1-P(R)+P(RnS)
=> 2P(S) = 1-3/4 + (3/4)P(S)
=> P(S)(2-3/4) = 1/4
=> P(S) = (1/4)/(5/4) = 1/5

b) P(S'nR) = P(SuR) - P(S) = (1-P(S)) - P(S) = 4/5 - 1/5 = 3/5

c) P(S|R) = P(SnR)/P(R) = 0.75P(S)/0.75 = P(S) = 1/5

Quite involved. sorry.
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CalculusMan
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Since R and S are independent P(R) = P(R|S) = 3/4

Since R and S are independent then S′ and R′ are also independent, so
P(S′ ∩ R′) = P(S′) × P(R′)
P(S) = P(S′ ∩ R′)
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Sushmeist
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If you wouldn't mind I'm sure I should be able to see it but brain really has switched off. Will have a break after this question I think (for about a month!!:p: )
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Sushmeist
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Oooh, thanks guys!! Now it makes sense!! ... Time for a well earned (:rolleyes: ) break!
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Christophicus
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Full solution above if you've missed it.
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Sushmeist
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Yep got it now thanks!!!
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