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# Good explanation needed for this one Jane has to pick 3

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Founder
Joined: 04 Dec 2002
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Kudos [?]: 28476 [0], given: 5105

Location: United States (WA)
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Good explanation needed for this one Jane has to pick 3 [#permalink]

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28 Jul 2003, 19:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?
тАв 0.2
тАв 0.25
тАв 0.4
тАв 0.8
тАв 0.96

Kudos [?]: 28476 [0], given: 5105

GMAT Instructor
Joined: 07 Jul 2003
Posts: 769

Kudos [?]: 234 [0], given: 0

Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Re: Probability 2 [#permalink]

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28 Jul 2003, 20:59
bb wrote:
Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?
тАв 0.2
тАв 0.25
тАв 0.4
тАв 0.8
тАв 0.96

Here are 3 ways of looking at this:
(1)
WE are already given that E is chosen, so the problem is simply what is the prob that she will pick B if she chooses 2 subjects.

This is the probability that she will either choose the subject as the first pick OR the subject as the second pick.

Pr(b 1st) = 1/5
Pr(b 2nd) = Pr(b not 1st AND b 2nd) = 4/5 * 1/4 = 1/5

These 2 are mutually exclusive, hence the probability is 1/5 + 1/5 = 2/5 = 0.4

Method 2:
The probability that she will NOT leave B as one of her 3 unpicked choice is exactly the same problem.
Pr(B is NOT one of 3 unpicked choices) = 4/5 * 3/4 * 3/2 = 2/5 = 0.4

Method 3:
Pr (b picked in 2 choices) = 1 - Pr(B not picked of her two choices)
= 1 - (4/5 * 3/4) = 1 - 3/5 = 2/5 = 0.4.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Kudos [?]: 234 [0], given: 0

Manager
Joined: 24 Jun 2003
Posts: 145

Kudos [?]: 3 [0], given: 0

Location: India
Re: Probability 2 [#permalink]

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29 Jul 2003, 06:54
bb wrote:
Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?
тАв 0.2
тАв 0.25
тАв 0.4
тАв 0.8
тАв 0.96

Another way of looking at this problem:

E is already chosen. Now B can be chosen in 2nd choice or third choice.

Number of ways B can come in second choice = 1C1*4C1 = 4
Number of ways B can come in third choice = 4C1*1C1 = 4

Total number of ways in which two subjects can be chosen, one by one, out of the remaining five = 5C1*4C1 = 20

Therefore, the probability that B will be chosen = 8/20 = 0.4

Kudos [?]: 3 [0], given: 0

SVP
Joined: 30 Oct 2003
Posts: 1788

Kudos [?]: 112 [0], given: 0

Location: NewJersey USA

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23 Jan 2004, 08:19
A different approach. Assume he has chosen E and B then only one other subject has to be chosen in 4C1 ways
Total ways to to choose 2 subjects from 5 is 5C2 = 10 ways

P = 4C1/5C2 = 0.4

Kudos [?]: 112 [0], given: 0

23 Jan 2004, 08:19
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# Good explanation needed for this one Jane has to pick 3

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