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Jim and John are workers in a department that has a total of six emplo

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Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 09 Oct 2015, 01:46
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57% (00:52) correct 43% (00:48) wrong based on 308 sessions

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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

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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 09 Oct 2015, 08:36
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Bunuel wrote:
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

Kudos for a correct solution.

Using probability rules, we get:

P(Jim and John BOTH chosen) = P(Either Jim or John is chosen first AND the other is chosen second)
= P(Either Jim or John is chosen first) x P(the other is chosen second)
= 2/6 x 1/5
= 1/15
= D

Aside: For the first probability, there are 2 out of 6 ways to select on of the two men (Jim and John).
For the second probability, there are now 5 people remaining and only 1 of the men (Jim or John) remains


For more on rewriting probabilities with AND and OR, see our free video: http://www.gmatprepnow.com/module/gmat- ... /video/754

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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 09 Oct 2015, 02:14
Probability = 2c2/6c2
= 1/15

or using probablity of individual events x number of arrangements possible = (1/6)x(1/5)x 2 =1/15

Answer D
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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 13 Oct 2015, 22:49
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jim is chosen first and then john is chosen = (1/6)*(1/5) or john is chosen first and then jim is chosen = (1/6)*(1/5)

either case is possible; prob = (1/6)*(1/5) + (1/6)*(1/5) = 1/15
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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 14 Oct 2015, 01:05
P(Jim)=1/6 and P(john)=1/5
P(both)=P(Jim)*P(john)*arrangements=1/6*1/5* 2(ways)

Ans=D (1/15)
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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 11 Nov 2016, 10:53
Bunuel wrote:
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

Kudos for a correct solution.


2C2/6C2 = 1/15

Answer will be (D) 1/15

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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 28 Sep 2017, 02:52
Probability = 2c2/6c2
= 1/15
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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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New post 03 Oct 2017, 08:39
Bunuel wrote:
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


The probability that both Jim and John are chosen is 2/6 x 1/5 = 2/30 = 1/15.

Alternate Solution:

The total number of groups of two employees out of six is 6C2 = 6!/[2!(6-2)!] = 6 x 5/2 x 1 = 15. Picking both Jim and John is one of those 15 choices, so the probability of choosing both of them is 1/15.

Answer: D
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Re: Jim and John are workers in a department that has a total of six emplo  [#permalink]

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