Probability: Boy or Girl?

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Probability: Boy or Girl?

by joconnor » Mon Dec 19, 2011 11:17 pm
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a) 3/8
b) 1/4
c) 3/16
d) 1/8
e) 1/16

Thanks!

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by ankush123251 » Mon Dec 19, 2011 11:26 pm
Each child has 2 options being a boy or a girl.
So total number of outcomes = 2* 2 * 2 * 2 =16.

Number of ways in which 4 children will have 2 boys and 2 girls = 4!/2!* 2! = 6.

So,
Probability = 6/16 =3/8

QA is A.

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by [email protected] » Tue Dec 20, 2011 12:08 am
joconnor wrote:A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a) 3/8
b) 1/4
c) 3/16
d) 1/8
e) 1/16

Thanks!
Total no. of outcomes = 2^4
Possibility of 2 girls and 2 boys = 4!/2!*2! = 6
Therefore, required probability = 6/2^4 = [spoiler]3/8[/spoiler]

The correct answer is A.
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by bubbliiiiiiii » Wed Dec 21, 2011 5:48 am
Hello Sir,

Could you please elaborate this statement?
Possibility of 2 girls and 2 boys = 4!/2!*2! = 6
I think the statement goes like this total permutations/[Number of ways in which two girls can be arranged * Number of ways in which 2 boys can be arranged].

Can you please post some more logic and how to solve such type of questions?
Regards,

Pranay

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by k.pankaj.r » Wed Dec 21, 2011 6:52 am
probability of having a boy = 1/2
probability of having a girl= 1/2

so probability of having 2 boys and 2 girls is (1/2)^4
but it be arranged in 4!/2!2!, since we have 2 identical cases i.e 2 boys and 2 girls.
so final answer is (1/16)X(4!/2!2!) which comes out be 3/8.
hence A

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by vishal.pathak » Wed Dec 28, 2011 11:56 am
[email protected] wrote:
joconnor wrote:A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a) 3/8
b) 1/4
c) 3/16
d) 1/8
e) 1/16

Thanks!
Total no. of outcomes = 2^4
Possibility of 2 girls and 2 boys = 4!/2!*2! = 6
Therefore, required probability = 6/2^4 = [spoiler]3/8[/spoiler]

The correct answer is A.
Hi Anurag,

Please explain the error in my approach

The couple can have

1. 4B,0G
2. 3B,1G
3. 2B,2G
4. 1B,3G
5. 0B,4G

So we have one favorable case out of these 5. So probability is 1/5

Regards,
Vishal

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by pemdas » Wed Dec 28, 2011 12:53 pm
hey Vishal, i think once you've asked this question on the forum or may be someone else has asked the same question. I remember this kind of query coming on from a member

anyways, you pre-arranged the sets here by saying 1 boy and three girls or two boys and two girls

these are not total possible outcomes but rather the sets pre-arranged by you

Plz compare children A,B,C,D may be borne with two independent gender type - male or female

when you say two boys and two girls it's not one outcome, but this is one event

the no of outcomes for this event is 6, two boys and two girls can be formed AB, CD, AD, AC, BD, BC

the same way all your events one boy and three girls -> no of outcomes would be 4C3

all four boys or girls 4C4

anyways, you need to understand the difference between event and outcome here

quit boys and girls, consider each child as a coin and it may land tail/head when tossed - probability 1/2 (independence of events)
in any combination the set with probabilities (yes will be boy and no will not be boy) is 1/2 by 1/2 and so on. Hence 1/2*1/2*1/2*1/2 or (1/2)^4

How many outcomes are possible for two boys or two girls, select among four children 4C2 or 6 ways

6*(1/4)^4=3/8
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by ArunangsuSahu » Fri Dec 30, 2011 9:26 am
Sample Space = 2^4( as each one can be either a boy or a girl)=16

Exactly 2 boys and exactly 2 girls in 4C2*2C2 ways = Event = 6

Required Probability = 6/16 = 3/8

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Re: Probability: Boy or Girl?

by [email protected] » Tue Apr 19, 2022 5:34 am
joconnor wrote:
Mon Dec 19, 2011 11:17 pm
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a) 3/8
b) 1/4
c) 3/16
d) 1/8
e) 1/16

Thanks!
We can solve this question using counting methods.
P(exactly 2 girls and 2 boys) = (number of 4-baby outcomes with exactly 2 girls and 2 boys)/(TOTAL number of 4-baby outcomes)

As always, we'll begin with the denominator.

TOTAL number of 4-baby arrangements
There are 2 ways to have the first baby (boy or girl)
There are 2 ways to have the second baby (boy or girl)
There are 2 ways to have the third baby (boy or girl)
There are 2 ways to have the fourth baby (boy or girl)
By the Fundamental Counting Principle (FCP), the total number of 4-baby arrangements = (2)(2)(2)(2) = 16

Number of 4-baby outcomes with exactly 2 girls and 2 boys
This portion of the question boils down to "In how many different ways can we arrange 2 G's and 2 B's (where each G represents a girl, and each B represents a boy)?"

----------ASIDE-------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
------------BACK TO THE QUESTION---------------------------

Our goal is to arrange the letters G, G, B, and B
There are 4 letters in total
There are 2 identical G's
There are 2 identical B's
So, the total number of possible arrangements = 4!/[(2!)(2!)] = 6

So.....
P(exactly 2 girls and 2 boys) = 6/16 = 3/8