Permutaion

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SwatiDenre: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Mon Apr 18, 2011 4:34 pm

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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Source: Beat The GMAT — Problem Solving | Fri Jun 10, 2011 12:50 pm

jainnikhil02: Master | Next Rank: 500 Posts; Posts: 123; Joined: Tue May 31, 2011 12:26 amFollowed by: 1 member; Location: Hyderabad; Thanked: 5 times

by jainnikhil02 » Fri Jun 10, 2011 1:10 pm

IMO A

Let me know if i am wrong

Nikhil K Jain
____________________

"Life is all about timing" Don't waste your and others time.

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SwatiDenre: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Mon Apr 18, 2011 4:34 pm

by SwatiDenre » Fri Jun 10, 2011 1:46 pm

Ans - B

But how did u get through it.

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

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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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

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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jainnikhil02: Master | Next Rank: 500 Posts; Posts: 123; Joined: Tue May 31, 2011 12:26 amFollowed by: 1 member; Location: Hyderabad; Thanked: 5 times

by jainnikhil02 » Fri Jun 10, 2011 8:56 pm

approch was rite but done calculation mistakte..

Nikhil K Jain
____________________

"Life is all about timing" Don't waste your and others time.

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