Problem Solving

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

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

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

[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