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
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Anurag@Gurome
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Jack and Jill work at a hospital with 4 other workers, so there are 6 workers. out of the 6, 2 to be chosen.
the probability of both jack and jill be chosen = the probability of jack chosen first and jill chosen next + the probability of jill chosen first and jack chosen next
the probability of jack chosen first = 1/6
the probability of jill chosen next in the 5 left = 1/5
the probability of jack chosen first and jill chosen next = the probability of jack chosen first * the probability of jill chosen next in the 5 left = 1/6 * 1/5
similarly, the probability of jill chosen first and jack chosen next = 1/6 * 1/5
the probability of both jack and jill be chosen = 2 * 1/6 * 1/5 = 1/15
hence, it is C
the probability of both jack and jill be chosen = the probability of jack chosen first and jill chosen next + the probability of jill chosen first and jack chosen next
the probability of jack chosen first = 1/6
the probability of jill chosen next in the 5 left = 1/5
the probability of jack chosen first and jill chosen next = the probability of jack chosen first * the probability of jill chosen next in the 5 left = 1/6 * 1/5
similarly, the probability of jill chosen first and jack chosen next = 1/6 * 1/5
the probability of both jack and jill be chosen = 2 * 1/6 * 1/5 = 1/15
hence, it is C
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another way of doing this is:
you need to have one selection (jack, jill) out of number of selections possible of 2 people in 6 people group
number of ways of selecting two people in 6 people = 6C2
jack and jill selection constitutes of 1 selection.
so the probability of selecting jack and jill = 1/ 6c2 = 1/15
you need to have one selection (jack, jill) out of number of selections possible of 2 people in 6 people group
number of ways of selecting two people in 6 people = 6C2
jack and jill selection constitutes of 1 selection.
so the probability of selecting jack and jill = 1/ 6c2 = 1/15
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GMATGuruNY
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P(Jack or Jill on the first pick) = 2/6. (Of the 6 people, either Jack or Jill.)vishugogo wrote: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
P(Jack or Jill on the second pick) = 1/5. (Of the 5 remaining people, either Jack or Jill, whoever wasn't the first pick.)
Since we want both events to happen, we multiply:
2/6 * 1/5 = 1/15.
The correct answer is C.
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vishugogo
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thank you
GMATGuruNY wrote:P(Jack or Jill on the first pick) = 2/6. (Of the 6 people, either Jack or Jill.)vishugogo wrote: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
P(Jack or Jill on the second pick) = 1/5. (Of the 5 remaining people, either Jack or Jill, whoever wasn't the first pick.)
Since we want both events to happen, we multiply:
2/6 * 1/5 = 1/15.
The correct answer is C.
