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conditional probability 2

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conditional probability 2

by sana.noor » Sat Mar 02, 2013 9:25 am
Alfonso is conducting a survey of families with 3 children. If a family is selected at random, what is the probability that the family will have exactly 2 boys if the second child is a boy?
a)3/8
b)4/8
c)1/2
d)1/4
e)2/8

C
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by Brent@GMATPrepNow » Sat Mar 02, 2013 10:00 am
sana.noor wrote:Alfonso is conducting a survey of families with 3 children. If a family is selected at random, what is the probability that the family will have exactly 2 boys if the second child is a boy?
a)3/8
b)4/8
c)1/2
d)1/4
e)2/8

C
There are other (faster) ways to solve this question, but if you're looking for a foolproof method, I suggest a probability tree.

I'll assume that the probability of having a boy = the probability of having a girl = 0.5
Since each outcome is equally likely, let's draw a tree diagram to see all of the possible outcomes:
Image

So, there are 8 possible outcomes and each outcome is equally likely.

Now it's given that the second child is a boy.
So, of the 8 possible outcomes, 4 of them are such that the second child is a boy.
Image


Now that we know we've narrowed the field to 4 equally-likely outcomes, we want the probability of having exactly 2 boys
Image
There are 2 such outcomes (out of 4 possible outcomes)
So, the probability equals [spoiler]2/4 = 1/2 = C[/spoiler]

Cheers,
Brent
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by sana.noor » Sat Mar 02, 2013 10:19 am
Thanku Thanku Thanku soooo much Brent...:)
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by Anurag@Gurome » Sat Mar 02, 2013 7:26 pm
sana.noor wrote:Alfonso is conducting a survey of families with 3 children. If a family is selected at random, what is the probability that the family will have exactly 2 boys if the second child is a boy?
We have three positions 1st, 2nd, and 3rd. For each position we have two choices, either boy or girl. But we know that the second position is already filled by a boy. Hence, total number of possibility for filling the positions = (2 choices for 1st position)*(2 choices for 2nd position) = 2*2 = 4

Now, among these 4 possibilities, how many cases are are there such that exactly two positions are filled by a boy?

We already know that the 2nd position is filled by a boy.
Hence, only one of other two positions can be filled by a boy.
There are 2 possibilities for such a case.

Hence, required probability = 2/4 = 1/2

The correct answer is C.
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by vipulgoyal » Sun Apr 07, 2013 3:43 am
if the second child is a boy

2*1*2/8 = 1/2 since order dosent matters
two boys
1*1*2/8 = 1/4 order doesnt matters
conditional probiblity
1/4 divide by 1/2 = 1/2 Ans
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by bharat.bondalapati » Sun Apr 07, 2013 8:57 am
_ _

We need B G or G B combination.

Therefore, (1/2)*(1/2) + (1/2)*(1/2) = 1/2
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by GMATGuruNY » Sun Apr 07, 2013 7:05 pm
sana.noor wrote:Alfonso is conducting a survey of families with 3 children. If a family is selected at random, what is the probability that the family will have exactly 2 boys if the second child is a boy?
a)3/8
b)4/8
c)1/2
d)1/4
e)2/8

C
Since good outcomes are BBG and GBB, what matters here is the gender of the THIRD child.
If the first child is a boy, then the third child must be a girl.
If the first child is a girl, then the third child must be a boy.
In other words, the gender of the third child must be DIFFERENT from that of the first child.
P(gender of the third child is different from that of the first child) = 1/2.

The correct answer is C.
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by freyesinsb » Tue Apr 09, 2013 11:43 am
Are answer choices really constructed like this on the GMAT??? B is equal to C and D is equal to E. Will I encounter this on test day? Normally, I would just say that A is right without trying to solve the problem just by looking at the possible answers.
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by Anju@Gurome » Tue Apr 09, 2013 11:59 am
freyesinsb wrote:Are answer choices really constructed like this on the GMAT??? B is equal to C and D is equal to E. Will I encounter this on test day? Normally, I would just say that A is right without trying to solve the problem just by looking at the possible answers.
Very good observation.
You'll never see such options in proper GMAT problems.
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