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
Permutaion
This topic has expert replies
-
- Junior | Next Rank: 30 Posts
- Posts: 13
- Joined: Mon Apr 18, 2011 4:34 pm
- jainnikhil02
- Master | Next Rank: 500 Posts
- Posts: 123
- Joined: Tue May 31, 2011 12:26 am
- Location: Hyderabad
- Thanked: 5 times
- Followed by:1 members
IMO A
Let me know if i am wrong
Let me know if i am wrong
Nikhil K Jain
____________________
"Life is all about timing" Don't waste your and others time.
____________________
"Life is all about timing" Don't waste your and others time.
-
- Junior | Next Rank: 30 Posts
- Posts: 13
- Joined: Mon Apr 18, 2011 4:34 pm
-
- Legendary Member
- Posts: 1085
- Joined: Fri Apr 15, 2011 2:33 pm
- Thanked: 158 times
- Followed by:21 members
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
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
Success doesn't come overnight!
- cans
- Legendary Member
- Posts: 1309
- Joined: Mon Apr 04, 2011 5:34 am
- Location: India
- Thanked: 310 times
- Followed by:123 members
- GMAT Score:750
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
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
If my post helped you- let me know by pushing the thanks button
Contact me about long distance tutoring!
[email protected]
Cans!!
Contact me about long distance tutoring!
[email protected]
Cans!!
- jainnikhil02
- Master | Next Rank: 500 Posts
- Posts: 123
- Joined: Tue May 31, 2011 12:26 am
- Location: Hyderabad
- Thanked: 5 times
- Followed by:1 members
approch was rite but done calculation mistakte..
Nikhil K Jain
____________________
"Life is all about timing" Don't waste your and others time.
____________________
"Life is all about timing" Don't waste your and others time.