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Probability: Official Prep

by tanvis1120 » Sat Aug 09, 2014 6:38 am
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

Answer:

Total number of ways in which boys and girls (4 children) can be born = 4! = 24
Now, the probability of a child being a girl or a boy is equally likely, that is, 1/2.
I cannot proceed further.
Please explain the solution to me by any method other than listing out the ways (BBGG, BGBG etc) manually.

Thanks in Advance!
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by GMATGuruNY » Sat Aug 09, 2014 7:26 am
A couple decides to have 4 children. lf 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(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get an equal number of boys and girls is for first two children born to be boys and for the last two children born to be girls.
P(BBGG) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16.

Total possible ways:
BBGG is only ONE WAY to get exactly 2 boys and exactly 2 girls.
Now we must account for ALL OF THE WAYS to get exactly 2 boys and exactly 2 girls.
Any arrangement of the letters BBGG will yield exactly 2 boys and 2 girls.
Thus, to account for ALL OF THE WAYS to get exactly 2 boys and 2 girls, the result above must be multiplied by the number of ways to arrange the letters BBGG.
Number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical B's and by another 2! to account for the two identical G's:
4!/(2!2!) = 6.

Multiplying the results above, we get:
P(exactly 2 boys and 2 girls) = 6 * 1/16 = 3/8.

The correct answer is A.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html
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by Brent@GMATPrepNow » Sat Aug 09, 2014 7:43 am
tanvis1120 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
An alternative to listing every possible outcome is to draw a a tree diagram. This will allow you to quickly see all of the outcomes.

Image

As you can see, all 16 possible scenarios are listed here.
Of these 16 cases, 6 of them meet the "2 boys and 2 girls" criterion.

So, the probability = [spoiler]6/16 = 3/8 = A[/spoiler]

Cheers,
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by Brent@GMATPrepNow » Sat Aug 09, 2014 7:51 am
tanvis1120 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
Another approach:

TOTAL NUMBER OF OUTCOMES:
2 possibilities (boy or girl) for the 1st child, 2 possibilities for the 2nd child, 2 possibilities for the 3rd child, and 2 possibilities for the 4th child.
By the Fundamental Counting Principle (FCP), the TOTAL number of outcomes = (2)(2)(2)(2) = 16

NUMBER OF OUTCOMES WHERE EXACTLY 2 CHILDREN ARE GIRLS
Say we have 4 generic children, and we want to choose 2 of them to be girls (the two remaining children will be the boys).
Since the order of the selected children doesn't matter, we can use combinations.
We can select 2 of the 4 children in 4C2 ways = 6 ways

So, P(2 boys and 2 girls) = 6/16 = [spoiler]3/8 = A[/spoiler]

Aside: If anyone is interested, we have a free video on calculating combinations (like 4C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789


Cheers,
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by abhasjha » Sat Aug 09, 2014 9:51 am
Dear Mitch and Brent ,

I was wondering if this question is exactly similar to the question - "if a coin is tossed 4 times then what is the probability of getting exactly two heads provided that the coin is fair (thereby meaning probability of heads and tails are equal?"

if that is so then we can straight way apply the formula ncrp^r q^(n-r). = 4c2(1/2)^2. (1/2)^2.

this is equal to 3/8 . hence answer A .
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by Brent@GMATPrepNow » Sat Aug 09, 2014 12:04 pm
Your coin question is exactly the same as the boy/girl question because P(boy) = P(girl) = P(heads) = P(tails) = 1/2

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by [email protected] » Sat Aug 09, 2014 5:33 pm
Hi tanvis1120,

Both Mitch and Brent have provided the "math" explanation that you were looking for, so I won't rehash that here.

I'm not sure if you were just interested in the how the math "worked" or if you were fundamentally against the idea of listing out all of the possibilities to answer the question. If it's the latter option, then you should know that sometimes the fastest way to get to the correct answer is by using "brute force" (in this case, listing out all of the possibilities).

75 minutes is enough time to get to all of the questions in the Quant section, but it's NOT enough time to take the "long approach" on every question. In many cases, the "math approach" IS the "long approach." While your practice should absolutely include some work on all the math (formulas, rules, etc.) that you need to know, it should also include lots of alternative approaches to questions.

To maximize your score on the GMAT, you have to be a flexible thinker, so keep an open mind to the occasional use of "brute force" - it can make some really tough questions really easy to answer.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by abhasjha » Sat Aug 09, 2014 10:10 pm
Thank you Brent.
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by tanvis1120 » Sun Aug 10, 2014 8:05 am
Thank you all for the explanations !
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by j_shreyans » Sat May 16, 2015 8:45 pm
Hi ,

We can have below possibilities for the boys and girls.

B1,B2,B3,B4,
B1,B2,B3,G
B1,B2,G1,G2
B,G1,G2,G3
G1,G2,G3,G4

so out of the 5 we have only one possibility for 2 boys and 2 girls

Total outcome is 4c2 = 6 so why not 1/6?

Please advise and correct me

Thanks
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by DavidG@VeritasPrep » Sun May 17, 2015 5:16 am
This approach will be somewhat time-consuming. You're off to is a good start, but we don't have the entire sample space yet. Consider B1,B2,B3,G , for example. It's possible that the girl in this family could be the oldest or the youngest, or somewhere in the middle, so we have to account for these scenarios too:

G,B,B,B,
B,G,B,B
B,B,G,B,
B,B,B,G

Similarly, If there's one boy, we've got to reason the same way:
B,G,G,G
G,B,G,G
G,G,B,G
G,G,G,B

And if there's two of each, we'd have:
B,B,G,G
B,G,B,G
B,G,G,B
G,B,G,B
G,G,B,B
G,B,B,G

And, of course, we've got
G,G,G,G
B,B,B,B

In sum: you've got 4 ways to have 1 girl, 4 ways to have 1 boy, 6 ways to have 2 boys and 2 girls, 1 way to have all boys, and 1 way to have all girls.
Add 'em up, and you've got 16. Because there are 6 ways to have 2 boys and 2 girls, the answer is 6/16 = 3/8. Whew. So you can see that both Brent and Mitch provide much better time-efficient ways to solve.
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by theCEO » Sun May 17, 2015 5:22 am
j_shreyans wrote:Hi ,

We can have below possibilities for the boys and girls.

B1,B2,B3,B4,
B1,B2,B3,G
B1,B2,G1,G2
B,G1,G2,G3
G1,G2,G3,G4

so out of the 5 we have only one possibility for 2 boys and 2 girls

Total outcome is 4c2 = 6 so why not 1/6?

Please advise and correct me

Thanks
You have listed 4 possibilities that can be expanded some more,
for example B1,B2,B3,G could also have a possibility of B1,G,B2,B3

Possibility of having all boys or all girls
1 + 1 = 2
B1,B2,B3,B4
G1,G2,G3,G4

Possibility of having 1 boy 3 girls or 3 boys 1 girl
4 x 2 = 8

B1,B2,B3,G
B1,B2,G,B3
B1,G,B2,B3
G,B1,B2,B3
swop the letters B and G and you will get the other 4 choices

Possibility of having 2 boys 2 girls
6
B1,B2,G1,G2
B1,G1,B2,G2
B1,G1,G2,B2
G1,B1,B2,G2
G1,G2,B1,B2,
G1,B1,G2,B2

Total possibilities are 2+8+6 = 16
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by Amrabdelnaby » Wed Nov 25, 2015 3:11 pm
Hi Brent,

I tend to find myself solving most of the probability questions, if not all, using counting techniques.

I do know that both of them are shadows of each other.

I just wanted to check with you, is it actually possible to rely solely on counting to solve probability?

thanks
GMATGuruNY wrote:
A couple decides to have 4 children. lf 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(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get an equal number of boys and girls is for first two children born to be boys and for the last two children born to be girls.
P(BBGG) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16.

Total possible ways:
BBGG is only ONE WAY to get exactly 2 boys and exactly 2 girls.
Now we must account for ALL OF THE WAYS to get exactly 2 boys and exactly 2 girls.
Any arrangement of the letters BBGG will yield exactly 2 boys and 2 girls.
Thus, to account for ALL OF THE WAYS to get exactly 2 boys and 2 girls, the result above must be multiplied by the number of ways to arrange the letters BBGG.
Number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical B's and by another 2! to account for the two identical G's:
4!/(2!2!) = 6.

Multiplying the results above, we get:
P(exactly 2 boys and 2 girls) = 6 * 1/16 = 3/8.

The correct answer is A.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html
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by Brent@GMATPrepNow » Wed Nov 25, 2015 3:33 pm
Amrabdelnaby wrote:Hi Brent,

I tend to find myself solving most of the probability questions, if not all, using counting techniques.

I do know that both of them are shadows of each other.

I just wanted to check with you, is it actually possible to rely solely on counting to solve probability?

thanks
No. There are some probability questions that can't be answered via counting techniques alone.
Here's one example - https://www.beatthegmat.com/independent- ... 87754.html
There are plenty of others.

Also, there will be times when using counting techniques will be MUCH SLOWER than applying probability rules.

Cheers,
Brent
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by Matt@VeritasPrep » Fri Nov 27, 2015 12:38 am
Amrabdelnaby wrote: I tend to find myself solving most of the probability questions, if not all, using counting techniques.

I do know that both of them are shadows of each other.

I just wanted to check with you, is it actually possible to rely solely on counting to solve probability?
Counting and probability are intimately related on many GMAT probability problems, but this relationship doesn't hold for too long. (For instance, suppose I asked you "If I pick a random letter from A to J, replace it, then pick a random letter from A to Z, what's the probability that the second comes earlier in the alphabet than the first?" This has a very simple solution, but counting won't help much!)

Counting is worth considering when (i) the probability involves order and (ii) each order has the same probability, or at least uses the same probability values, but don't become too dependent on it.
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