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Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N)

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Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N) [#permalink] New post   11 Sep 2020, 11:45
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Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N)=3/4. Find P(N).

A) 1/12

B) 1/4

C) 5/12

D) 5/6

E) 1/2

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C
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Re: Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N) [#permalink] New post   11 Sep 2020, 15:31
If the events M and N are not mutually exclusive then:
P(M or N)= P(M) + P(N) - P(M and N)
3/4 = 1/2 + P(N) - 1/6
P(N)=5/12
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Re: Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N) [#permalink] New post   11 Sep 2020, 15:42
I don't know why but the sign is wrong, anyway:

P(M)+P(MandN)+P(MorN)+P(N)=1
0.5+0.75+0.17+P(N)=1
P(N)=-0.42 -> (5/12)

Answer C
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Re: Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N) [#permalink] New post   11 Sep 2020, 21:05
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Given:

P(M) = \(\frac{1}{2}\)

P(M or N) = P(M U N) = \(\frac{3}{4}\)

P(M and N) = P(M \(\cap\) N) = \(\frac{1}{6}\)

By definition, P(M U N) = P(M) + P(N) - P(M \(\cap\) N)


\(\frac{3}{4} = \frac{1}{6} + P(N) - \frac{1}{6}\)


\(P(N) = \frac{3}{4} + \frac{1}{6} - \frac{1}{2} = \frac{9 + 2 - 6}{12} = \frac{5}{12}\)


Option C

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Re: Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N) [#permalink] New post   12 Sep 2020, 00:30
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P(M or N) = P(M) + P(N) - P(M and N)

=> \(\frac{3}{4}\) = \(\frac{1}{2}\) + P(N) - \(\frac{1}{6}\)

=> P(N) = \(\frac{3}{4}\) + \(\frac{1}{6}\) - \(\frac{1}{2}\)

=> P(N) = \(\frac{(9 + 2 - 6) }{ 12}\)

=> P(N) = \(\frac{5}{12}\)

Answer C
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Re: Let M and N be events such that P(M)=1/2, P(M and N)=1/6 and P(M or N) [#permalink]
12 Sep 2020, 00:30
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