Jack and Jill work at a hospital with 4 other workers. For a

Question

Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3

nave · Accepted Answer

Total number of people = 6

Probability of selecting Jack first and then Jill is  \frac{1}{6}  *\frac{1}{5}  =\frac{1}{30} 
Probability of selecting Jill first and then Jack is  \frac{1}{6}  *\frac{1}{5}  =\frac{1}{30}

Therefore probability of selecting Jack and Jill for the review is  \frac{1}{30}  +\frac{1}{30}  =\frac{1}{15}

Answer is C

Bunuel · Answer

1/6C2=1/15.

Answer: C.

redfield · Answer

I just did chance of 1 being selected first time is 1/6 + chance of 1 being selected second time 1/5, turns into 2/30 and reduces to 1/15. Is that the same as 1/6C2=1/15 (my math fundamentals are still lacking a bit)?

ENGRTOMBA2018 · Answer

Yes. nCr = n!/(r!*(n-r)!), thus 6C2 = 6!/(2!*4!) = 15

Thus 1/6C2 = 1/15

The correct way to approach is a follows:

Selecting 1 out of the 2 = 2/6 and then the remaining 1 person out of Jack and Jill = 1/5 , giving you the final probability = 2/6*1/5 = 1/15

Hope this helps.

Bunuel · Answer

Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3

Divyadisha · Answer

Total number of ways to choose 2 out of 6 workers= 6!/2!4!= 15

Number of ways to choose both jack and jill= 1

probability= 1/15

C should be the answer

Skywalker18 · Answer

Probability that Jack and Jill will both be chosen out of 6 workers 
= (2/6)*(1/5)
= 1/15
Answer C

abhishekdadarwal2009 · Answer

Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3

IMO C

Sol: Prob of Jack being selected=1/6 then the prob of jill being selected=1/5 (as only 5 are left for second selection)
then the other way jill selected=1/6 and jack selected=1/5

so we get (1/6*1/5)+(1/6*1/5)=(1/30)*2=1/5

Basshead · Answer

Number of ways to choose 2 from 6: 6C2 = 15 ways

Number of ways to choose Jack and Jill from 2: 2C2 = 1 way

= 1/15

Another approach:

Probability of choosing Jack first: 2/6
Probability of choosing Jill second: 1/5

1/6 * 2/5 = 2/30 = 1/15

bumpbot · Answer

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