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# A couple decides to have 4 children. If they succeed in havi

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A couple decides to have 4 children. If they succeed in havi  [#permalink]

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13 Mar 2014, 02:29
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Difficulty:

75% (hard)

Question Stats:

52% (01:44) correct 48% (01:36) wrong based on 1766 sessions

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

Problem Solving
Question: 160
Category: Arithmetic Probability
Page: 83
Difficulty: 700

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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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13 Mar 2014, 02:29
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SOLUTION

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

$$P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}$$, we should mulitply by $$\frac{4!}{2!2!}$$ as the scenario GGBB can occur in several ways: GGBB, BBGG, GBGB, ... # of scenarios possible is # of permutations of 4 letters GGBB out of which 2 G's and 2 B's are identical and equals to $$\frac{4!}{2!2!}$$.

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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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25 Apr 2014, 01:06
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16
no of ways of getting P(GGBB) is 4!/2!*2!;
Total no of ways is 2^n =2^4 =16;

6/16 = 3/8;
We can consider this question to a coin that is flipped for 4 times . what is the probability of getting exactly two heads .

P(all out comes) = 1/2 *1/2 *1/2 *1/2 =1/16;

P(favorable outcomes) = 4!/(2! * 2!) = 6/16 =3/8;

(OR)

Second Approach

GBGB
GGBB
BBGG
BGBG
GBBG
BGGB

6 possible ways .
total no of ways is
Baby can be a boy or a girl.
For each baby the probability is 1/2 ;for 4 babies it's 1/16;
6/16 = 3/8;

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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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15 Mar 2014, 10:59
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Re: A couple decides to have 4 children. If they succeed in having 4 child  [#permalink]

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10 Aug 2014, 02:40
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By fundamental counting principle,
Total No of out comes: 2^4 = 16

Total Desired outcomes: No of ways to arrange BBGG by MISSISSIPPI rule= 4!/(2!*2!) = 6

Probability is 6/16 = 3/8
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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Updated on: 15 Mar 2015, 00:07
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4
So here we have four positions and there are two options to fill each position
so total number of cases=2x2x2x2=16
now we need 2boys and 2 girls OR we can say that we simply need 2 boys because if its not a boy it has to be a girl
favourable cases=4C2=6
probability=6/16=3/8

(A)
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Originally posted by masoomdon on 14 Mar 2015, 01:30.
Last edited by masoomdon on 15 Mar 2015, 00:07, edited 1 time in total.
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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14 Mar 2015, 08:14
3
1/2 is the chance of boy or girl

1/2 * 1/2 * 1/2 * 1/2 = 1/16 (4 children, boy or girl)

Possible ways of 2 boys, 2 girls:

GGBB
BBGG
GBGB
BGBG
BGGB
GBBG

= 6 ways

we need "OR" --> 1/16 OR 1/16 ....
1/16 + 1/16 + 1/16 ....... +1/16 = 6/16 = 3/8

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Re: A couple decides to have 4 children. If they succeed in having 4 child  [#permalink]

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16 Jul 2015, 20:06
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery.
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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03 Nov 2015, 21:17
1
2
Binomial Probability
The probability of achieving exactly k successes in n trials is shown below.
Formula: P(Probability of K successes in n trials) = nCk p^k q^n-k

n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial

According to our question

n(4 children) = 4
k( we want exactly 2 girls) = 2
n – k = 2
p (probability of getting a girl in one trial) = 1/2
q = 1 – p = 1/2

4C2 * 1/2^2 * 1/2^2 =3/8

Binomial can also be used for problem like coins.

Binomial is used when following conditions are satisfied.

Fixed number of trials
Independent trials
Two different classifications
Probability of success stays the same for all trials
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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04 Nov 2015, 01:08
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

Let us handle a single case. First boy, second boy, third girl, fourth girl. The probability of this case is 1/2*1/2*1/2*1/2 = 1/16
Two boys or two girls can be born among 4 children in 4C2 ways= 6 ways
So the required probability is: total number of ways * probability of one such way= 6* (1/16)=3/8
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Re: A couple decides to have 4 children. If they succeed in having 4 child  [#permalink]

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04 Feb 2016, 16:50
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery.
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?

Bump seeking an answer to the question above:

I got 1/5 as well.

BBBB
BBBG
BBGG
GGGB
GGGG

1/5 outcomes

The question doesn't say anything about the order. How do we know when order matters?
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Re: A couple decides to have 4 children. If they succeed in having 4 child  [#permalink]

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04 Feb 2016, 22:35
4
1
Dondarrion wrote:
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery.
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?

Bump seeking an answer to the question above:

I got 1/5 as well.

BBBB
BBBG
BBGG
GGGB
GGGG

1/5 outcomes

The question doesn't say anything about the order. How do we know when order matters?

In this case we are saying that the probability of BGGG is the same as the probability of BBGG. But that is not so.

You can have 1 boy and 3 girls in 4 ways: BGGG, GBGG, GGBG, GGGB
But you can have 2 boys and 2 girls in 6 ways: BBGG, BGGB, GGBB, BGBG, GBGB, GBBG

So the probability depends on the number of ways in which you can get 2 boys and 2 girls.

Think of it this way: if you throw two dice, is the probability of getting a sum of 2 same as the probability of getting a sum of 8? No.
For sum fo 2, you must get 1 + 1 only.
For sum of 8, you could get 4 + 4 or 3 + 5 or 2 + 6 etc.
So probability of getting sum of 8 would be higher.

In the same way, the order matters in this question.
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Re: A couple decides to have 4 children. If they succeed in having 4 child  [#permalink]

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06 Aug 2016, 15:17
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Re: A couple decides to have 4 children. If they succeed in having 4 child  [#permalink]

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08 Aug 2016, 03:22
1
1
ManonZ wrote:

Do you mean in the total number of cases? If yes, then it is actually 2^4.
First child can happen in 2 ways (boy or girl).
Second, third and fourth kids can also happen in 2 ways.
Total number of ways = 2*2*2*2 = 16
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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23 Feb 2018, 13:22
1
Hi All,

There are a couple of ways to answer this question:

1) You could make a table of all the options (there are only 16 possible outcomes) and count up all the ways to get 2 boys and 2 girls

or

2) You can do the math

Here's the math approach:

Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities

It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question...

4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls

6/16 = 3/8 = A

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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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18 Mar 2018, 11:27
How do you arrive at this being a "600 level difficulty" problem? It's the 8th from the last problem in the OG 18 Supplemental Quant Guide.

I believe it'd be considered at least a 700 level difficulty by GMAC, but I could be wrong.

Could you explain how you determined it to be 600 level difficulty?

Thanks, Bunuel! You're the best
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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18 Mar 2018, 12:07
destinyawaits wrote:
How do you arrive at this being a "600 level difficulty" problem? It's the 8th from the last problem in the OG 18 Supplemental Quant Guide.

I believe it'd be considered at least a 700 level difficulty by GMAC, but I could be wrong.

Could you explain how you determined it to be 600 level difficulty?

Thanks, Bunuel! You're the best

You can check difficulty level of a question along with the stats on it in the first post. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. For this particular question it's 700-level.
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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07 May 2018, 10:28
3
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

We need to determine the probability of P(B-B-G-G).

P(B-B-G-G) = (1/2)^4 = 1/16

The number of ways to arrange B-B-G-G is 4C2 = 4!/(2! x 2!) = 3 x 2 = 6.

Thus, the total probability is 6/16 = 3/8.

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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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04 Jan 2019, 12:50
I thought that order does not matter for when child is born, therefore
Total number of events = 5 (explained below)

0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl

Probability (2 B and 2 G exactly) = 1/5

Not sure where I am assuming and making a mistake.
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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04 Jan 2019, 14:01
1
shavarna wrote:
I thought that order does not matter for when child is born, therefore
Total number of events = 5 (explained below)

0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl

Probability (2 B and 2 G exactly) = 1/5

Not sure where I am assuming and making a mistake.

Hi shavarna,

The 5 options you've listed are NOT all equally likely, so you would have to do a bit more work to get to the correct answer. There are a couple of ways to answer this question:

1) You could make a table of all the options (since each child could be a boy or a girl, there are only 2^4 = 16 possible outcomes) and determine all the ways to get 2 boys and 2 girls

or

2) You can do the math

Here's the math approach:

Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities. It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question...

4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls out of 16 possibilities. 6/16 = 3/8

GMAT assassins aren't born, they're made,
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Re: A couple decides to have 4 children. If they succeed in havi   [#permalink] 04 Jan 2019, 14:01

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