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probability doubt : Need More explanation

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probability doubt : Need More explanation

by bitsho » Sat Jul 24, 2010 1:05 pm
Ques : A person has 3 children with atleast one boy . Find the probability of having atleast 2 boys among children (Source : Arun Sharma , Quant Apt for CAT, Problem:14.5)

Solution:

The event is occuring under the following situations :

event 1 :2nd is a boy & 3rd is a girl OR
event 2 : 2nd is a girl and 3rd is a boy OR
event 3 :2nd is a boy & 3rd is a boy

i.e. required prob = 1/2*1/2+1/2*1/2*+1/2*1/2 = 3/4

However,I am unable to understand why we are calculating the prob for event 1 (2nd is a boy & 3rd is a girl ) & event 2 (2nd is a girl and 3rd is a boy) . In my opinion , both represent the same . since the ques is asking about prob of atleast 2 boys , how does the order of girl-boy came in to pic . As per my understanding , prob should be

event 1 :2nd is a boy & 3rd is a girl OR
event 2 : 2nd is a boy & 3rd is a boy

i.e. required prob = 1/2*1/2+1/2*1/2 = 1/2

Please throw some light where i am missing the point .

Any help is appreciated
Last edited by bitsho on Sat Jul 24, 2010 2:21 pm, edited 1 time in total.
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by akdayal » Sat Jul 24, 2010 1:57 pm
Ques : A person has 3 children with atleast one boy . Find the probability of having atleast 2 boys among children (Source : Arun Sharma , Quant Apt for CAT, Problem:14.5)
I also have doubt on answer.
According to question
Total number of ways/events: 1B2G, 2B1G, 3B0G ( As it mentioned in question that person has 3 children with atleast 1 B)
Hence No of fav events: 2B1G, 3B0G (atleast 2B among children)

Hence req. prob = 2/3

I do not think order will matter in this case. If it is can anyone tell me why?
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by GMATGuruNY » Sat Jul 24, 2010 3:02 pm
bitsho wrote:Ques : A person has 3 children with atleast one boy . Find the probability of having atleast 2 boys among children (Source : Arun Sharma , Quant Apt for CAT, Problem:14.5)

Solution:

The event is occuring under the following situations :

event 1 :2nd is a boy & 3rd is a girl OR
event 2 : 2nd is a girl and 3rd is a boy OR
event 3 :2nd is a boy & 3rd is a boy

i.e. required prob = 1/2*1/2+1/2*1/2*+1/2*1/2 = 3/4

However,I am unable to understand why we are calculating the prob for event 1 (2nd is a boy & 3rd is a girl ) & event 2 (2nd is a girl and 3rd is a boy) . In my opinion , both represent the same . since the ques is asking about prob of atleast 2 boys , how does the order of girl-boy came in to pic . As per my understanding , prob should be

event 1 :2nd is a boy & 3rd is a girl OR
event 2 : 2nd is a boy & 3rd is a boy

i.e. required prob = 1/2*1/2+1/2*1/2 = 1/2

Please throw some light where i am missing the point .

Any help is appreciated
Remember:

P(good outcome) = 1 - P(bad outcome)

It's given that we have at least one boy. The question is really asking:

What is the probability that at least 1 of the other 2 children is a boy?

The only bad outcome is if the two other children are girls: 1/2 * 1/2 = 1/4.
So P(at least one additional boy) = 1 - 1/4 = 3/4.

Also, remember that:

P(A and B) = P(A) * P(B)

To determine the probability of events happening together, we have to determine the probability of each separate event and multiply -- and we have to account for all the different ways those separate events could occur.

So P(at least 1B) = P(B) * P(G) + P(G) * P(B) + P(B) * P(B)

Of course, it's much simpler to use the method I described above.
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by bitsho » Sat Jul 24, 2010 3:33 pm
Thanks Mitch ! thanks a tonne . so the take away from this problem is

we have to account for all the different ways those separate events could occur.

Thanks again for clarifying this point
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by Stuart@KaplanGMAT » Sat Jul 24, 2010 8:00 pm
bitsho wrote:Ques : A person has 3 children with atleast one boy . Find the probability of having atleast 2 boys among children (Source : Arun Sharma , Quant Apt for CAT, Problem:14.5)
Hi,

where does it say that there's an equal chance that any child is a boy or girl? All the solutions assume that there's a 50/50 chance, but it doesn't say so explicitly.

Without that information, the question is unanswerable.
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by bitsho » Sun Jul 25, 2010 2:40 am
Thanks Stuart ! you have pointed out a basic flaw in question :) I do agree with your point
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