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Re: From a group of 8 volunteers, including Andrew and Karen, 4 [#permalink]
GreenlightTestPrep wrote:
Carcass wrote:
From a group of 8 volunteers, including Andrew and Karen, 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


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)

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).

the video below shows you how to calculate combinations (like 6C3) in your head

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)
= 2/7

Answer: D

Cheers,
Brent



Hi Brent,

Shouldnt probability of choosing Andrew be 1/8?How come we are assuming it to be 1?

From my understanding,it should be 1/8 x 6c3 divided by 8c4.Not sure how we are taking "1" for Andrew's selection
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 [#permalink]
vaishar3 wrote:

Hi Brent,

Shouldnt probability of choosing Andrew be 1/8?How come we are assuming it to be 1?

From my understanding,it should be 1/8 x 6c3 divided by 8c4.Not sure how we are taking "1" for Andrew's selection


I had the same concern - it's not guaranteed that Andrew will be selected, right?
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 [#permalink]
3
jaidevphadke wrote:
vaishar3 wrote:

Hi Brent,

Shouldnt probability of choosing Andrew be 1/8?How come we are assuming it to be 1?

From my understanding,it should be 1/8 x 6c3 divided by 8c4.Not sure how we are taking "1" for Andrew's selection


I had the same concern - it's not guaranteed that Andrew will be selected, right?



You cannot keep one thing probabilistic and one thing fixed. If you take probability of choosing Andrew, then you need to multiply by the probability of not choosing Karen. And the probabilty of not choosing Karen is not 6c3. The probability of not choosing Karen is 6/7 * 5/6 * 4/5. Therefore your total probability comes to 1/14.

But then you assumed that Andrew will be the first person to be selected. What if he is the last person to be selected? Therefore you need to multiply it by 4, since Andrew can be selected in any one of the 4 slots. So total is 1/14 * 4 = 2/7
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 [#permalink]
2
(1) The ratio Andrew is selected in the end = 4/8 = 1/2

(2) The ratio Karen is selected= 3/7
(3) -> The ration Karen is not selected = 1-3/7 = 4/7

(1) x (3) = 1/2 * 4/7 = 2/7

So answer is (D)
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 [#permalink]
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