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From a group of 8 volunteers, including karen and Andrew, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
A) 3/7 B) 5/12 C) 27/70 D)2/7 E)9/35
I am sorry if this question has been covered in another post, but here is my take on the problem posted above.
Andrew has a 50% percent chance of being chosen, as there are a total of 8 volunteers and 4 will be chosen at random. Then, we can view the second part of the question as an complimentary question (or something not happening)
Therefore the probability of Karen being chosen is 3/7. Then... 1-3/7 (the complimentary) is 4/7.
We can the take 1/2 x 4/7 to be 4/14. 4/14 simplified is 2/7 Answer D
Could somebody please clarify whether this is an acceptable method for the question, or whether I achieved the corrected answer simply out of luck. The OA is D
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Total # selections possible: 8C4 = 70 Andrew has his spot fix so we need to select for remaining 3 spots Since Karen should not be in the team, we have 6 people to select from for 3 spots= 6C3 = 20
From a group of 8 volunteers, including karen and Andrew, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
A) 3/7 B) 5/12 C) 27/70 D)2/7 E)9/35
I am sorry if this question has been covered in another post, but here is my take on the problem posted above.
Andrew has a 50% percent chance of being chosen, as there are a total of 8 volunteers and 4 will be chosen at random. Then, we can view the second part of the question as an complimentary question (or something not happening)
Therefore the probability of Karen being chosen is 3/7. Then... 1-3/7 (the complimentary) is 4/7.
We can the take 1/2 x 4/7 to be 4/14. 4/14 simplified is 2/7 Answer D
Could somebody please clarify whether this is an acceptable method for the question, or whether I achieved the corrected answer simply out of luck. The OA is D
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.