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06 Jun 2021, 08:49
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Hello,
I have a doubt in the conceptual understanding of probability.
Following question has 2 methods to solve the same:
Q- A class has 4 girls (4G) and 5 boys(5B). 3 kids are randomly selected. What is the probability that exactly 1 girl is selected?
Method1- Probability of selecting GBB= 4/9 x 5/8 x 4/7 = 10/63
Now, there are 3!/2! = 3 ways of arranging 1G and 2B
So, final probability = 3 x (10/63)= 10/21
Method2-
Probability (1G and 2B)= (no. of ways of selecting 1G and 2B)/ (total no. of ways of selecting 3 kids from 9)
For numerator,
Selecting 1G from 4G = 4C1= 4 ways
Selecting 2B from 5B= 5C2 = 10 ways
So, total ways to select 1G and 2B= 4 x 10= 40 ways
For Denominator,
total ways to select 3 kids from 9 kids= 9C3= 84 ways
So, Probability (1G and 2B)= 40/84= 10/21
Now, in method 1, we consider the order of GBB by multiplying probability of (GBB) by 3!/2!= 3
However, in method 2, the calculation doesn't accommodate the order of GBB it seems.
Can any math expert please help me with this? What am I missing here?
Bunuel, VeritasKarishma, egmat, GMATCoachBen
I have a doubt in the conceptual understanding of probability.
Following question has 2 methods to solve the same:
Q- A class has 4 girls (4G) and 5 boys(5B). 3 kids are randomly selected. What is the probability that exactly 1 girl is selected?
Method1- Probability of selecting GBB= 4/9 x 5/8 x 4/7 = 10/63
Now, there are 3!/2! = 3 ways of arranging 1G and 2B
So, final probability = 3 x (10/63)= 10/21
Method2-
Probability (1G and 2B)= (no. of ways of selecting 1G and 2B)/ (total no. of ways of selecting 3 kids from 9)
For numerator,
Selecting 1G from 4G = 4C1= 4 ways
Selecting 2B from 5B= 5C2 = 10 ways
So, total ways to select 1G and 2B= 4 x 10= 40 ways
For Denominator,
total ways to select 3 kids from 9 kids= 9C3= 84 ways
So, Probability (1G and 2B)= 40/84= 10/21
Now, in method 1, we consider the order of GBB by multiplying probability of (GBB) by 3!/2!= 3
However, in method 2, the calculation doesn't accommodate the order of GBB it seems.
Can any math expert please help me with this? What am I missing here?
Bunuel, VeritasKarishma, egmat, GMATCoachBen
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06 Jun 2021, 10:23
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An indirect answer: why does method 1 need to account for the ordering?
Method 1
10/63 = the probability of the specific ordering: G G B
But the question doesn’t ask for the probability of G first, G second, and B third. In fact, the probability of that particular order could be called the “specific probability”.
The question asks about all different formations of G G B, of which there are three.
............
Method 2
When dealing with combinations, ordering doesn’t matter because it’s already anticipated (in a sense).
Applying the combination formula to choosing 2 boys (call them A and B) out of 5 anticipates that the boys could come in the order of A B and B A. Either way, it doesn’t matter.
Very subtle for sure. I hope I was clear. Feel free to ask any questions of course.
Posted from my mobile device
Method 1
10/63 = the probability of the specific ordering: G G B
But the question doesn’t ask for the probability of G first, G second, and B third. In fact, the probability of that particular order could be called the “specific probability”.
The question asks about all different formations of G G B, of which there are three.
............
Method 2
When dealing with combinations, ordering doesn’t matter because it’s already anticipated (in a sense).
Applying the combination formula to choosing 2 boys (call them A and B) out of 5 anticipates that the boys could come in the order of A B and B A. Either way, it doesn’t matter.
Very subtle for sure. I hope I was clear. Feel free to ask any questions of course.
Posted from my mobile device
06 Jun 2021, 19:41
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In a lot of moderately complicated probability questions, you have the option to solve in two different ways, and the question you ask about is a good example:
- you can pretend you're making each selection one at a time, in order. So here, we can pretend we're picking a first child, a second child, and a third child. To get exactly one girl, we might pick G, B, B, in that order. The probability we do that is (4/9)(5/8)(4/7). But that's not the only way to get exactly one girl. We might also pick B, G, B, or we might pick B, B, G. The probability either of those things happen will be identical to the first case, because there's no reason picking the girl last will be any more or less likely than picking her first, so the answer is 3 * (4/9)(5/8)(4/7).
- or you can pretend you're making all of the selections simultaneously, so order does not matter. So here, to work out how many selections are possible in total, we can work out how many sets of 3 children can be chosen from 9 when order does not matter. That's precisely what "9C3" means (if order matters when picking 3 things from 9, you have 9*8*7 possibilities, but if order is irrelevant, you have 9C3 or 9*8*7/3! possibilities). If we're going to solve this way, we need to be sure order also does not matter when we count the possibilities for our numerator. And we have 5C2 ways of picking 2 boys when their order does not matter, and 4C1 ways of picking the one girl, and we can now multiply our choices from each group to find out how many possible selections there are in total of one girl and two boys, and that gives us our numerator. So the right answer is also (4C1)(5C2)/9C3.
For a lot of questions, either approach works almost equally well, and you should just choose whichever one feels more natural to you (I prefer the first approach for conceptual reasons, but people who have practiced the second a lot will often find it easier, and there's no particularly compelling reason to prefer one to the other). The one caveat, regardless of which approach you choose, is that you must be consistent: if you decide to count your options for the denominator as if the order of your selections doesn't matter, you must be absolutely sure, when you count your options for the numerator, that order also does not matter for those selections. If you pretend order matters for the denominator, say, but doesn't for the numerator, your answer will always be wildly off.
Posted from my mobile device
- you can pretend you're making each selection one at a time, in order. So here, we can pretend we're picking a first child, a second child, and a third child. To get exactly one girl, we might pick G, B, B, in that order. The probability we do that is (4/9)(5/8)(4/7). But that's not the only way to get exactly one girl. We might also pick B, G, B, or we might pick B, B, G. The probability either of those things happen will be identical to the first case, because there's no reason picking the girl last will be any more or less likely than picking her first, so the answer is 3 * (4/9)(5/8)(4/7).
- or you can pretend you're making all of the selections simultaneously, so order does not matter. So here, to work out how many selections are possible in total, we can work out how many sets of 3 children can be chosen from 9 when order does not matter. That's precisely what "9C3" means (if order matters when picking 3 things from 9, you have 9*8*7 possibilities, but if order is irrelevant, you have 9C3 or 9*8*7/3! possibilities). If we're going to solve this way, we need to be sure order also does not matter when we count the possibilities for our numerator. And we have 5C2 ways of picking 2 boys when their order does not matter, and 4C1 ways of picking the one girl, and we can now multiply our choices from each group to find out how many possible selections there are in total of one girl and two boys, and that gives us our numerator. So the right answer is also (4C1)(5C2)/9C3.
For a lot of questions, either approach works almost equally well, and you should just choose whichever one feels more natural to you (I prefer the first approach for conceptual reasons, but people who have practiced the second a lot will often find it easier, and there's no particularly compelling reason to prefer one to the other). The one caveat, regardless of which approach you choose, is that you must be consistent: if you decide to count your options for the denominator as if the order of your selections doesn't matter, you must be absolutely sure, when you count your options for the numerator, that order also does not matter for those selections. If you pretend order matters for the denominator, say, but doesn't for the numerator, your answer will always be wildly off.
Posted from my mobile device
