SDW2 wrote:
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,
GMATCoachBenYes, this is a tricky thing to explain and understand. That is the reason I am adding my bit here inspite of perfect explanations above.
To know whether "order matters" or not, think this way:
Probability of GBB= 4/9 x 5/8 x 4/7 = 10/63
What is GBB? It is the probability of selecting a girl, then a boy, and then a boy (notice your reducing denominators so you are selecting them one after the other)
Now is it what is asked? Is it "the probability that exactly 1 girl is selected"? No. This is the probability of selecting first a girl, then a boy and then another boy. So what do we do?
There are two more ways of selecting exactly one girl - BGB and BBG. So you need to adjust for them too.
BGB = 5/9 x 4/8 x 4/7 = 10/63 (notice that the order of the numerators here is different though essentially they give the same probability)
BBG = 5/9 x 4/8 x 4/7 = 10/63 (again, the order of the numerators here is different though essentially they give the same probability)
So we simply multiply by 3 instead of calculating each of GBB, BGB and BBG. We are NOT arranging the girl and the boys since we are not required to. We are calculating the different ways in which we can pick one girl and two boys. Just that both the things are equal numerically.
Since it is just a selection, the simple combinations approach works too.
Even if we were to arrange after selecting, in probability the answer would still be the same though the question doesn't ask us to.