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Good explanation needed for this one Jane has to pick 3

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Good explanation needed for this one Jane has to pick 3 [#permalink] New post 28 Jul 2003, 18:54
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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
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Re: Probability 2 [#permalink] New post 28 Jul 2003, 19:59
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.
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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

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Re: Probability 2 [#permalink] New post 29 Jul 2003, 05:54
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
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 [#permalink] New post 23 Jan 2004, 07:19
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
  [#permalink] 23 Jan 2004, 07:19
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Good explanation needed for this one Jane has to pick 3

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