From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?
A 3/7
B 5/12
C 27/70
D 2/7
E 9/35
Probability question - Adam selected and not Karen
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I think the question is missing some key information.mdecarbo wrote:From a group of volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?
A 3/7
B 5/12
C 27/70
D 2/7
E 9/35
How many volunteers are there altogether?
Cheers,
Brent
Hey Brent - thanks for answering the questions. Long time no talk, I'm back on the GMAT grind - taking it in 2 weeks. I did accidentally leave out that there are 8 volunteers, by which 4 will be selected random for the charity group.
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Hey Mike,
I hope you're doing well.
You might want to go back and edit your first post, so that others will see the complete question.
Cheers,
Brent
I hope you're doing well.
You might want to go back and edit your first post, so that others will see the complete question.
Cheers,
Brent
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P(Andrew is selected but Karen is not selected) = (number of 4-person groups with Andrew but not Karen)/(total # of 4-person groups possible)mdecarbo wrote:From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?
A 3/7
B 5/12
C 27/70
D 2/7
E 9/35
number of 4-person groups with Andrew but not Karen
Take the task of creating groups and break it into stages.
Stage 1: Place Andrew in the 4-person group
We can complete this stage in 1 way
Stage 2: Send Karen out of the room and, from the remaining 6 volunteers, select 3 more people.
Since the order in which we select the 3 volunteers does not matter, we can use combinations.
We can select 3 people from 6 volunteers in 6C3 ways (20 ways).
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 4-person group) in (1)(20) ways (= 20 ways)
total # of 4-person groups possible
We can select 4 people from all 8 volunteers in 8C4 ways ( = 70 ways).
So, P(Andrew is selected but Karen is not selected) = (20)/(70)
= [spoiler]2/7 = D[/spoiler]
------------------------------
Cheers,
Brent
Aside: If anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
Hi Brent,Brent@GMATPrepNow wrote: Send Karen out of the room and, from the remaining 6 volunteers, select 3 more people.
Since the order in which we select the 3 volunteers does not matter, we can use combinations.
We can select 3 people from 6 volunteers in 6C3 ways (20 ways).
In case if we have to choose 4 people include Karen, the desired out come will be 6C2 = 15. Is that right?
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Exactly.feedrom wrote:Hi Brent,Brent@GMATPrepNow wrote: Send Karen out of the room and, from the remaining 6 volunteers, select 3 more people.
Since the order in which we select the 3 volunteers does not matter, we can use combinations.
We can select 3 people from 6 volunteers in 6C3 ways (20 ways).
In case if we have to choose 4 people include Karen, the desired out come will be 6C2 = 15. Is that right?
We automatically place Karen and Adam on the "team," and then select 2 more people from the remaining 6 people.
Cheers,
Brent
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Hi Brent!!
Could you please also give an alternate solution where order matters. I'm getting confused when trying to solve the other way.
Or what is wrong with my approach:
# of ways we may select remaining three persons= 6 x 5 x 4
# of ways we may select all four out of eight = 8 x 7 x 6 x 5
Probablity of required selection = (6 x 5 x 4)/(8 x 7 x 6 x 5)= 1/14
Thank you!
Could you please also give an alternate solution where order matters. I'm getting confused when trying to solve the other way.
Or what is wrong with my approach:
# of ways we may select remaining three persons= 6 x 5 x 4
# of ways we may select all four out of eight = 8 x 7 x 6 x 5
Probablity of required selection = (6 x 5 x 4)/(8 x 7 x 6 x 5)= 1/14
Thank you!
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From the 8 people, 4 will be selected.mdecarbo wrote:From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?
A 3/7
B 5/12
C 27/70
D 2/7
E 9/35
Thus, P(A is selected) = 4/8.
From the 7 remaining people, 4 will NOT be selected.
Thus, P(K is not selected) = 4/7.
To combine these probabilities, we multiply:
4/8 * 4/7 = 2/7.
The correct answer is D.
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Perfect!feedrom wrote:It's very impressive, Mitch!
Could I calculate like this:
Probability =( Adam, not Karen, not Karen, not Karen) x 4 (arrangements) = 1/8*6/7*5/6*4/5*4 =2/7 ?
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You almost have it.parveen110 wrote:Hi Brent!!
Could you please also give an alternate solution where order matters. I'm getting confused when trying to solve the other way.
Or what is wrong with my approach:
# of ways we may select remaining three persons= 6 x 5 x 4
# of ways we may select all four out of eight = 8 x 7 x 6 x 5
Probability of required selection = (6 x 5 x 4)/(8 x 7 x 6 x 5)= 1/14
Thank you!
If we're saying that order matters, we need to remember that Adam can be selected 1st, 2nd, 3rd, or 4th.
So, for each of the (6 x 5 x 4) selections that include Adam, we can select Adam in 4 ways.
So, Probability of required selection = (6 x 5 x 4 x 4)/(8 x 7 x 6 x 5) = 2/7
Cheers,
Brent