A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
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A committee of 2 people is to be selected out of
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DavidG@VeritasPrep
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Yes! (You can think of probability as a ratio of combinations or permutations.)VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
P(no preachers) = # ways we can select committee with no preachers/# ways we can select committee total
# ways we can select no preachers: Both selections would be teachers. If there are 3 to choose from, we want 3C2 = 3*2/2! = 3.
# ways we can select committee total: There are 7 people. We want to, or 7C2 = 7*6/2! = 21
P(no preachers) = 3/21 = 1/7
1 - P(no preachers) = 1 - 1/7 = 6/7. The answer is D
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DavidG@VeritasPrep
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You can also go pick by pick. We want 1 - P( no preachers)VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
P(no preachers) = P(pick 1 not a preacher) * P(pick 2 not a preacher given that pick 1 was not.)
P(pick 1 not a preacher) = 3/7
P(pick 2 not a preacher given that pick 1 was not) = 2/6
P(pick 1 not a preacher) * P(pick 2 not a preacher given that pick 1 was not) = 3/7 * 2/6 = 6/42 = 1/7
1 - P(no preachers) = 1 - 1/7 = 6/7, or D
The gist is that you can think of probability as a ratio of combinations, or you can go pick by pick. (Just make sure that if you go pick by pick that you consider every possible order of selection.)
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Brent@GMATPrepNow
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Here's an approach that uses probability rules.VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
We want P(select at least 1 preacher)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A NOT happening)
So, here we get: P(getting at least 1 preacher) = 1 - P(NOT getting at least 1 preacher)
What does it mean to not get at least 1 preacher? It means getting zero preachers.
So, we can write: P(getting at least 1 preacher) = 1 - P(getting zero preachers)
P(getting zero preachers)
P(getting zero preachers) = P(1st selection is teacher AND 2nd selection is teacher)
= P(1st selection is teacher) x P(2nd selection is teacher)
= 3/7 x 2/6
= 1/7
So, P(getting at least 1 preacher) = 1 - 1/7
=6/7
Answer: D
Cheers,
Brent
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Vincen
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Hello Vjesus12.
This is how I would solve it: I'd use combinations and then probability.
There are 7 people and we have to pick 2, this can be done from $$7C2=\frac{7!}{5!2!}=\frac{7\cdot6\cdot5!}{5!\cdot2}=\frac{42}{2}=21.$$ Now:
The number of ways to pick 1 preacher is given by: $$3C1\cdot4C1=\frac{3!}{2!1!}\cdot\frac{4!}{3!1!}=3\cdot4=12.$$ The number of ways to pick 2 preachers is given by: $$4C2=\frac{4!}{2!2!}=\frac{12}{2}=6.$$ Hence, the total of ways to pick at least 1 preacher is 12+6 = 18.
Now, the probability of pick at least 1 preacher is: $$P=\frac{18}{21}=\frac{6}{7}.$$ Therefore, the correct answer is the option D.
This is how I would solve it: I'd use combinations and then probability.
There are 7 people and we have to pick 2, this can be done from $$7C2=\frac{7!}{5!2!}=\frac{7\cdot6\cdot5!}{5!\cdot2}=\frac{42}{2}=21.$$ Now:
The number of ways to pick 1 preacher is given by: $$3C1\cdot4C1=\frac{3!}{2!1!}\cdot\frac{4!}{3!1!}=3\cdot4=12.$$ The number of ways to pick 2 preachers is given by: $$4C2=\frac{4!}{2!2!}=\frac{12}{2}=6.$$ Hence, the total of ways to pick at least 1 preacher is 12+6 = 18.
Now, the probability of pick at least 1 preacher is: $$P=\frac{18}{21}=\frac{6}{7}.$$ Therefore, the correct answer is the option D.
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Scott@TargetTestPrep
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To determine how many ways to select at least one preacher, we can use the formula:VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
Number of ways to select at least 1 preacher = total number of ways to select the committee - number of ways to select NO preachers
Total number of ways to select the committee of 2 people is:
7C2 = 7!/[2!(7-2)!] = (7 x 6)/2! = 21
The number of ways to select NO preachers in the committee is:
3C2 = (3 x 2)/2! = 3
Thus, the number of ways to select at least 1 preacher is 21 - 3 = 18.
So the probability that the committee will be composed of at least 1 preacher is 18/21 = 6/7.
Alternate Solution:
We use the rule of the complement, which says: The probability of picking AT LEAST one preacher = 1 - the probability of picking NO preachers.
Note that the probability of picking NO preachers is the same as the probability of picking all teachers for this committee of 2. Thus, the probability of picking 2 teachers is 3/7 x 2/6 = 1/7.
Thus, the probability of picking at least one preacher = 1 - 1/7= 6/7.
Answer: D
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