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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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New post 13 Mar 2014, 02:29
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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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New post 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 \(\frac{4!}{2!2!}\).

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

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New post 25 Apr 2014, 01:06
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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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New post 14 Mar 2014, 04:48
Would Pascal's Triangle be a useful tool to quickly arrive at the answer in this case?
Using the triangle, I'm getting 3/8 as the answer, which I think is right, but I don't know if there are any limitations / specific cases in which Pascals Triangle should not be used for calculations like these. Please advise
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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New post 15 Mar 2014, 10:59
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21. Combinatorics/Counting Methods



For more:
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Hope it helps.
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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New post Updated on: 15 Mar 2015, 00:07
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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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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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New post 14 Mar 2015, 08:14
2
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

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

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New post 03 Nov 2015, 21:17
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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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New post 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 havi  [#permalink]

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New post 23 Feb 2018, 13:22
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

Final Answer =
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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New post 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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New post 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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New post 01 May 2018, 01:19
Bunuel wrote:
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 \(\frac{4!}{2!2!}\).

Answer: A.



hello friends:) pushpitkc, generis, niks18, Tuesday is announced as international day of Combinatorics and Probabilities, so yeah i decided why not to try some questions:)

can you guys please explan why my approach doesnt work here, for instance here it worked https://gmatclub.com/forum/there-are-10 ... l#p2053582 why wouldnt the same approach work here :? ?

here is my reasoning:

First of all i need to find all possible combinations

\(4C4 = 1\) ( all 4 are boys )

\(4C4 = 1\) (all 4 are girls )

\(4C1 * 4C3 = 4*4 =16\) (1 boy and 3 girls)

\(4C2 * 4C3 = 6*6 = 36\) (2 boys and 2 girls)

\(4C3*4C1 = 4*4 =16\) (3 boys and 1 girl )

Now from the above mentioned combinations i see that there are 36 ways to have 2 boys and 2 girls.

Total number of ways is \(1+1+16+36+16 = 70\)

So probability to have 2 boys and 2 girls is \(\frac{36}{70}\) :?

i think i have problem in applying correctly the concepts where we need to use both combinatorics and probability concept in one solution ... always get confused

For instance Bunuel`s solution includes probability, let me break it down his soluton.

\(P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}\),

So lets start with this \(\frac{4!}{2!2!}\) is it the same as \(\frac{4!}{2!(4-2)!}\) ? Or is it something like \(\frac{X!}{X!*Y!*Z!}\) i do remember that there is some formula in combinatorics when where there similar sorts of objects that have some different features... hard to explain when you have a vague idea :)

how about fraction \((\frac{1}{2})^4\) why is he mupltiplyng by \((\frac{1}{2})^4\) ...whats the logic ? :?

many thanks and have a great clear spring day :)
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A couple decides to have 4 children. If they succeed in havi  [#permalink]

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New post 01 May 2018, 02:49
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dave13 wrote:
Bunuel wrote:
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 \(\frac{4!}{2!2!}\).

Answer: A.



hello friends:) pushpitkc, generis, niks18, Tuesday is announced as international day of Combinatorics and Probabilities, so yeah i decided why not to try some questions:)

can you guys please explan why my approach doesnt work here, for instance here it worked https://gmatclub.com/forum/there-are-10 ... l#p2053582 why wouldnt the same approach work here :? ?

here is my reasoning:

First of all i need to find all possible combinations

\(4C4 = 1\) ( all 4 are boys )

\(4C4 = 1\) (all 4 are girls )

\(4C1 * 4C3 = 4*4 =16\) (1 boy and 3 girls)

\(4C2 * 4C3 = 6*6 = 36\) (2 boys and 2 girls)

\(4C3*4C1 = 4*4 =16\) (3 boys and 1 girl )

Now from the above-mentioned combinations i see that there are 36 ways to have 2 boys and 2 girls.

Total number of ways is \(1+1+16+36+16 = 70\)

So probability to have 2 boys and 2 girls is \(\frac{36}{70}\) :?

i think i have problem in applying correctly the concepts where we need to use both combinatorics and probability concept in one solution ... always get confused

For instance Bunuel`s solution includes probability, let me break it down his soluton.

\(P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}\),

So lets start with this \(\frac{4!}{2!2!}\) is it the same as \(\frac{4!}{2!(4-2)!}\) ? Or is it something like \(\frac{X!}{X!*Y!*Z!}\) i do remember that there is some formula in combinatorics when where there similar sorts of objects that have some different features... hard to explain when you have a vague idea :)

how about fraction \((\frac{1}{2})^4\) why is he mupltiplyng by \((\frac{1}{2})^4\) ...whats the logic ? :?

many thanks and have a great clear spring day :)



Hi dave13

Happy international day of Combinatorics and Probabilities :)

The previous example involved choosing from 4 books and 6 books respectively.
Here, we are required to choose from 4 kids. So, if 2 of the 4 kids are boys, the
other 2 kids need to be girls.

So, your combinations will be as follows:
\(4C4 = 1\) (all 4 are boys )
\(4C4 = 1\) (all 4 are girls )
\(4C3 * 1C1 = 4*1 = 4\) (1 boy and 3 girls)
\(4C2*2C2 = 6*1\) (2 boys and 2 girls)
\(4C3 * 1C1 = 4*1 = 4\) (3 boys and 1 girl)

The total possibilities are 1+1+4+6+4 = 16 and possibilities we have 2 boys and 2 girls are 6
Therefore, the probability that they will have exactly 2 girls and 2 boys is \(\frac{6}{16}\) or \(\frac{3}{8}\) (Option A)

Hope this clears your confusion!
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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New post 03 May 2018, 17:13
I approached this problem the same way Bunuel did.

GGBB = 4C2*2C2 = 6 ways of arranging the kids that order.

Equal chance of having a boy or girl so the probability for each is 1/2.

(1/2)^4 * 6 = 3/8

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

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New post 07 May 2018, 10:28
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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


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.

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

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New post 09 May 2018, 04:31
Bunuel wrote:
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 \(\frac{4!}{2!2!}\).

Answer: A.

Bunuel Pls explain (1/2)^4?
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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New post 09 May 2018, 09:43
siddreal wrote:
Bunuel wrote:
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 \(\frac{4!}{2!2!}\).

Answer: A.

Bunuel Pls explain (1/2)^4?


2 girls and 2 boys (4 children) each having the probability of 1/2, so 1/2*1/2*1/2*1/2 = (1/2)^4.
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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New post 09 May 2018, 16:58
Bunuel wrote:
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 \(\frac{4!}{2!2!}\).

Answer: A.


Can you also consider this a combination question? 4C2 (boys) and 2C2 girls? Seem to see conflicting results below your solution that suggest both permutation/combination work.
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Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

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New post 10 May 2018, 02:22
Bunuel wrote:
siddreal wrote:
Bunuel wrote:
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 \(\frac{4!}{2!2!}\).

Answer: A.

Bunuel Pls explain (1/2)^4?


2 girls and 2 boys (4 children) each having the probability of 1/2, so 1/2*1/2*1/2*1/2 = (1/2)^4.


Bunuel I am sorry for not asking the question properly.
I understood the concept of (1/2)^4 but meant to ask why is 4!/2!2! not sufficient to solve the answer? Why do we have to add (1/2)^4 to the expression?
Kindly explain elaborately. :angel:
Re: A couple decides to have 4 children. If they succeed in havi &nbs [#permalink] 10 May 2018, 02:22

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