Probability: Official Prep

Expert replies

tanvis1120: Senior | Next Rank: 100 Posts; Posts: 95; Joined: Sat Dec 28, 2013 8:53 amFollowed by: 5 members; Location: United States; Thanked: 2 times

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!

Like this post

Source: Beat The GMAT — Problem Solving | Sat Aug 09, 2014 6:38 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Like this post

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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.

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,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Like this post

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Like this post

abhasjha: Master | Next Rank: 500 Posts; Posts: 379; Joined: Tue Sep 30, 2008 7:17 amFollowed by: 11 members; Location: NY; Thanked: 28 times

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 .

https://graceeducation1613.wordpress.co ... questions/

Like this post

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Like this post

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

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]

Like this post

abhasjha: Master | Next Rank: 500 Posts; Posts: 379; Joined: Tue Sep 30, 2008 7:17 amFollowed by: 11 members; Location: NY; Thanked: 28 times

by abhasjha » Sat Aug 09, 2014 10:10 pm

Thank you Brent.

https://graceeducation1613.wordpress.co ... questions/

Like this post

tanvis1120: Senior | Next Rank: 100 Posts; Posts: 95; Joined: Sat Dec 28, 2013 8:53 amFollowed by: 5 members; Location: United States; Thanked: 2 times

by tanvis1120 » Sun Aug 10, 2014 8:05 am

Thank you all for the explanations !

Like this post

j_shreyans: Legendary Member; Posts: 510; Joined: Thu Aug 07, 2014 2:24 amFollowed by: 5 members; Thanked: 3 times

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

Like this post

DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

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.

Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course

Like this post

theCEO: Master | Next Rank: 500 Posts; Posts: 363; Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members; Thanked: 115 times

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

Like this post

Amrabdelnaby: Master | Next Rank: 500 Posts; Posts: 137; Joined: Fri Nov 13, 2015 11:01 amFollowed by: 2 members; Thanked: 1 times

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

Like this post

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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

Brent Hanneson - Creator of GMATPrepNow.com

Like this post

Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

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.

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review