A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?
A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 1/2
The OA is C.
I'm really confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
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A committee of 3 has to be formed randomly...
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BTGmoderatorLU
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Hi LUANDATO,
We're told that a committee of 3 has to be formed randomly from a group of 6 people (including Tom and Mary). We're asked for the probability that Tom will be selected into the committee but Mary will not. This question can be solved by using the Combination Formula twice (in slightly different ways).
To start, we should calculate the total possible groups of 3... 6!/3!3! = (6)(5)(4)/(3)(2)(1) = 20 possible groups of 3
At this point, since there are so few possibilities, you could just list them all out. However, if you want to use Combination Formula to find the total number of groups that includes Tom but NOT Mary, here's how you would do it. We assign one of the 3 'spots' to Tom, meaning that there are 2 remaining spots (and neither of the two can go to Mary). This means that there are 4 possible people who can take those two spots... 4!/2!2! = (4)(3)/(2)(1) = 6 possible groups with Tom but NOT Mary
Probability of a group of 3 having Tom but NOT Mary = 6/20 = 3/10
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that a committee of 3 has to be formed randomly from a group of 6 people (including Tom and Mary). We're asked for the probability that Tom will be selected into the committee but Mary will not. This question can be solved by using the Combination Formula twice (in slightly different ways).
To start, we should calculate the total possible groups of 3... 6!/3!3! = (6)(5)(4)/(3)(2)(1) = 20 possible groups of 3
At this point, since there are so few possibilities, you could just list them all out. However, if you want to use Combination Formula to find the total number of groups that includes Tom but NOT Mary, here's how you would do it. We assign one of the 3 'spots' to Tom, meaning that there are 2 remaining spots (and neither of the two can go to Mary). This means that there are 4 possible people who can take those two spots... 4!/2!2! = (4)(3)/(2)(1) = 6 possible groups with Tom but NOT Mary
Probability of a group of 3 having Tom but NOT Mary = 6/20 = 3/10
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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GMATGuruNY
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From the 6 people, 3 will be selected.LUANDATO wrote:A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?
A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 1/2
Thus, P(Tom is selected) = 3/6 = 1/2.
From the 5 remaining people, 3 will NOT be selected.
Thus, P(Mary is not selected) = 3/5.
To combine these probabilities, we multiply:
1/2 * 3/5 = 3/10.
The correct answer is C.
Alternate approach:
ONE WAY:
One way to yield a favorable outcome is to choose Aaron first.
P(1st person selected is Aaron) = 1/6.
P(2nd person selected is not Karen) = 4/5. (Of the 5 remaining people, anyone but Karen.)
P(3rd person selected is not Karen) = 3/4. (Of the 4 remaining people, anyone but Karen.)
To combine these probabilities, we multiply:
1/6 * 4/5 * 3/4 = 1/10.
OTHER WAYS:
Since a favorable outcome will be achieved if Aaron is chosen 1st, 2nd, or 3rd -- for a total of 3 ways -- we multiply by 3:
1/10 * 3 = 3/10.
The correct answer is C.
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The total number of committees of 3 people from a group of 6 people that can be formed is 6C3 = (6 x 5 x 4)/(3 x 2) = 20. The number of committees can be formed in which Tom will be selected and Mary will not consists of Tom's being chosen for a spot, and 2 of 4 individuals (but not Mary) chosen for the remaining 2 spots:BTGmoderatorLU wrote:A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?
A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 1/2
The OA is C.
I'm really confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
1C1 x 1C0 x 4C2 = 1 x 1 x (4 x 3)/2 = 6
Therefore, the probability that Tom will be selected for the committee but Mary will not is 6/20 = 3/10.
Answer: C
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