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by SwatiDenre » Fri Jun 10, 2011 12:50 pm
Let A, B and C denotes separate generic traits. Suppose the probability that an individual will exhibit trait A is 1/2 and the probability that he exhibits B trait is 3/4 and the probability that he exhibits trait C is 3/5. What is the probability that he exhibits only one of the traits?

A. 1/4
B. 11/40
C. 1/6
D. 17/36
E. 11/36
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by jainnikhil02 » Fri Jun 10, 2011 1:10 pm
IMO A

Let me know if i am wrong
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by SwatiDenre » Fri Jun 10, 2011 1:46 pm
Ans - B

But how did u get through it.
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by pemdas » Fri Jun 10, 2011 2:28 pm
this one is mixed concept q., probability+permutation

P(not A)=1-1/2=1/2
P(not B)=1-3/4=1/4
P(not C)=1-3/5=2/5

P(only A)=P(A)*P(not B)*P(not C)=1/2*1/4*2/5=1/20
P(only B)=P(B)*P(not A)*P(not C)=3/4*1/2*2/5=3/20
P(only C)=P(C)*P(not A)*P(not B)=3/5*1/2*1/4=3/40

1/20+3/20+3/40=11/40
SwatiDenre wrote:Let A, B and C denotes separate generic traits. Suppose the probability that an individual will exhibit trait A is 1/2 and the probability that he exhibits B trait is 3/4 and the probability that he exhibits trait C is 3/5. What is the probability that he exhibits only one of the traits?

A. 1/4
B. 11/40
C. 1/6
D. 17/36
E. 11/36
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by cans » Fri Jun 10, 2011 8:11 pm
P(A)=1/2
p(B)=3/4
p(C)=3/5
p(one of A,B,C) = p(A & not B * not C) + p(B & not A * not C) + p(C & not B * not A)
= (1/2)(1/4)(2/5) + (3/4)(1/2)(2/5) + (3/5)(1/4)(1/2)
=11/40
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by jainnikhil02 » Fri Jun 10, 2011 8:56 pm
approch was rite but done calculation mistakte.. :(
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