I'm little confused about something here.
In the following questions my answers did not tally when i solved it using both combinations and probabii=lity methods.
the question is as follows:
1. Jim and John are workers in a department that has a total of six employees. Their boss decides that two workers from the department will be picked at random to participate in a company interview. What is the probability that both Jim and John are chosen?
A. 1/5
B. 1/6
C. 1/30
D. 1/15
E. 11/30
Combinations: (6C1 x 5C1)/6C2 = 1/15
probability: P(john & jean) = p(john) x P(jean given that john was selected) = 1/6 x 1/5 = 1/30
Can you please tell me what did i do wrong here?
Thanks
Probability v.s combinations
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Think in terms of total groupings.
The total number of ways that Jim and John can be selected is 1 > 2C2 if you are thinking about formulas.
Then, the total ways to people can be selected. 6C2 = 6 x 5 /2 = 15
1/15 is your answer
The total number of ways that Jim and John can be selected is 1 > 2C2 if you are thinking about formulas.
Then, the total ways to people can be selected. 6C2 = 6 x 5 /2 = 15
1/15 is your answer
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Hi Amrabdelnaby,
In your probability calculation, you assume that John has to be selected first, but that's not a requirement. John OR Jim could have been selected first... with the other selected second. THAT calculation is...
(2/6)(1/5) = 2/30 = 1/15
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
In your probability calculation, you assume that John has to be selected first, but that's not a requirement. John OR Jim could have been selected first... with the other selected second. THAT calculation is...
(2/6)(1/5) = 2/30 = 1/15
Final Answer: D
GMAT assassins aren't born, they're made,
Rich

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Hi Rich,
I guess I should have done the following if i wanted to do the probability rule.
1/6 x 1/5 x 2! (which is the number of ways to arrange to people) to give an answer of 1/15.
Thanks Rich for the clarification
I guess I should have done the following if i wanted to do the probability rule.
1/6 x 1/5 x 2! (which is the number of ways to arrange to people) to give an answer of 1/15.
Thanks Rich for the clarification
[email protected] wrote:Hi Amrabdelnaby,
In your probability calculation, you assume that John has to be selected first, but that's not a requirement. John OR Jim could have been selected first... with the other selected second. THAT calculation is...
(2/6)(1/5) = 2/30 = 1/15
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
GMAT/MBA Expert
 [email protected]
 Elite Legendary Member
 Posts: 10392
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HI Amrabdelnaby,
Yes, what you've described would certainly have gotten you the correct answer. The interesting thing about most GMAT questions (including Permutations, Combinations and Probability questions) is that they can be solved in a number of different ways. Ultimately, you're looking for an approach that will get you the correct answer, while still being efficient (so that you have enough time to get to all of the questions). As long as you have a reasonable 'shot' at all 37 Quant and all 41 Verbal questions, then you can approach the prompts however you see fit.
GMAT assassins aren't born, they're made,
Rich
Yes, what you've described would certainly have gotten you the correct answer. The interesting thing about most GMAT questions (including Permutations, Combinations and Probability questions) is that they can be solved in a number of different ways. Ultimately, you're looking for an approach that will get you the correct answer, while still being efficient (so that you have enough time to get to all of the questions). As long as you have a reasonable 'shot' at all 37 Quant and all 41 Verbal questions, then you can approach the prompts however you see fit.
GMAT assassins aren't born, they're made,
Rich