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21 Sep 2020, 17:15
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Hi hennaja,
If you've read through the thread of posts, you'll see that there are a number of different ways that you can answer this question. Depending on which method you prefer, you might find it easiest to treat this question as a Probability question (and calculate ALL the possible ways to get "what we want")….
We have 8 people to choose from; we'll pick 4 - we WANT Andrew in our group, we DON'T WANT Karen in our group.
_ _ _ _
If we want Andrew to be the first person we pick, then the other 3 CAN'T be Karen. We end up with the following math (notice that after picking each person, the total number of people remaining decreases):
(1/8)(6/7)(5/6)(4/5) = 4/56 = 1/14
Andrew COULD be the second, third or fourth person chosen though. Here's the math if he's SECOND and we don't pick Karen in any of the other spots...
(6/8)(1/7)(5/6)(4/5) = 4/56 = 1/14
Notice how the product is the SAME.
If he's third, it would still be 1/14
if he's fourth, it would still be 1/14
With ALL of those possibilities, we end up with 1/14 + 1/14 +1/14 + 1/14 = 4/14 = 2/7
GMAT assassins aren't born, they're made,
Rich
If you've read through the thread of posts, you'll see that there are a number of different ways that you can answer this question. Depending on which method you prefer, you might find it easiest to treat this question as a Probability question (and calculate ALL the possible ways to get "what we want")….
We have 8 people to choose from; we'll pick 4 - we WANT Andrew in our group, we DON'T WANT Karen in our group.
_ _ _ _
If we want Andrew to be the first person we pick, then the other 3 CAN'T be Karen. We end up with the following math (notice that after picking each person, the total number of people remaining decreases):
(1/8)(6/7)(5/6)(4/5) = 4/56 = 1/14
Andrew COULD be the second, third or fourth person chosen though. Here's the math if he's SECOND and we don't pick Karen in any of the other spots...
(6/8)(1/7)(5/6)(4/5) = 4/56 = 1/14
Notice how the product is the SAME.
If he's third, it would still be 1/14
if he's fourth, it would still be 1/14
With ALL of those possibilities, we end up with 1/14 + 1/14 +1/14 + 1/14 = 4/14 = 2/7
GMAT assassins aren't born, they're made,
Rich
14 Mar 2021, 13:45
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Bookmarks
With all the explanations given above, I am maybe wrong but I did this:
Probability that Andrew is selected : 4/8 = 1/2
If we now count the probability that Karen is not selected given that Andrew is: 4/7
1/2 * 4/7 = 4/14 = 2/7
Am I wrong ?
Bunuel EMPOWERgmatRichC
JeffTargetTestPrep
VeritasKarishma
MathRevolution
Probability that Andrew is selected : 4/8 = 1/2
If we now count the probability that Karen is not selected given that Andrew is: 4/7
1/2 * 4/7 = 4/14 = 2/7
Am I wrong ?
Bunuel EMPOWERgmatRichC
JeffTargetTestPrep
VeritasKarishma
MathRevolution
26 Mar 2021, 03:31
Kudos
Bookmarks
Moussaillon
It is correct.
Think of it this way - You make everyone stand randomly in a line and select first 4 and drop the rest. So probability that Andrew is in the first 4 is 4/8. Now one slot is gone.
Next, Karen should be in the last four so probability of that is 4/7.
Overall probability = (4/8)*(4/7) = 2/7
18 Feb 2022, 18:45
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KarishmaB
I'm not very familiar with probability & combinations.
Could you please check if my logic is correct here?
Big thanks!
Method 1. Probability + unarrange
1/8 (Andrew)
*
6/7*5/6*4/5 (Other 3 members excluding Karen)
*
4!/3!*1 (unarrange) (numerator: choose 4 people/ denominator: the order for the remaining members doesn't member
Method 2. Combination
1 * 6!/3!3! (Choose Andrew > Choose 3 members from the remaining 6. Their orders don't matter, so do the other 3 not selected)
_________
8!/4!4! ( Choose 4 members from the 8 in total. Their orders don't matter, so do the other 4 not selected)
I'm not very familiar with probability & combinations.
Could you please check if my logic is correct here?
Big thanks!
Method 1. Probability + unarrange
1/8 (Andrew)
*
6/7*5/6*4/5 (Other 3 members excluding Karen)
*
4!/3!*1 (unarrange) (numerator: choose 4 people/ denominator: the order for the remaining members doesn't member
Method 2. Combination
1 * 6!/3!3! (Choose Andrew > Choose 3 members from the remaining 6. Their orders don't matter, so do the other 3 not selected)
_________
8!/4!4! ( Choose 4 members from the 8 in total. Their orders don't matter, so do the other 4 not selected)
