Probability and combinations

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Probability and combinations

by viskhot » Thu Aug 29, 2013 8:32 pm
Q:

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3

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by Brent@GMATPrepNow » Thu Aug 29, 2013 8:50 pm
viskhot wrote: From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
One approach is to use counting techniques.

Probability = (# of ways to select 2 boys and 2 girls)/(total number of ways to select 4 children)

# of ways to select 2 boys and 2 girls
Take the task of selecting 2 boys and 2 girls, and break it into 2 stages:

Stage 1: Select 2 boys from 3 boys
Since the order of the selected boys does not matter, we can use combinations
We can select 2 boys from 3 boys in 3C2 ways (= 3 ways)

Stage 2: Select 2 girls from 3 girls
We can select 2 girls from 3 girls in 3C2 ways (= 3 ways)

By the Fundamental Counting Principle (FCP) we can complete both stages in (3)(9) ways 9 ways


Total number of ways to select 4 children
There are 6 children and we want to select 4 of them.
This can be accomplished in 6C4 ways ( = 6C2 ways = 15 ways)


So, the probability = 9/15 = [spoiler]3/5 = D[/spoiler]

Aside: If anyone is interested, we have a free video on calculating combinations (like 6C4) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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by ganeshrkamath » Fri Aug 30, 2013 12:02 am
viskhot wrote:Q:

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
4 children are selected - 2 boys and 2 girls
Favorable cases = 3C2 * 3C2 = 3 * 3 = 9

Sample space = 6C4 = 6C2 = 6*5/2 = 15

Required probability = 9/15 = 3/5

Choose D

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by vipulgoyal » Fri Aug 30, 2013 1:01 am
3/6*2/5*3/4*2/3 ways fo selecting 2 boys and two girls
now multiplied by no of ways by which these BBGG can be arranged among themself = 6
3/6*2/5*3/4*2/3*6=3/5

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by GMATGuruNY » Fri Aug 30, 2013 1:39 am
viskhot wrote:Q:

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
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 to select 2 boys, then 2 girls:
BBGG.
P(B on the 1st pick) = 3/6. (Of the 6 children, 3 are boys.)
P(B on the 2nd pick) = 2/5. (Of the 5 remaining children, 2 are boys.)
P(G on the 3rd pick) = 3/4. (Of the 4 remaining children, 3 are girls.)
P(G on the last pick) = 2/3. (Of the 3 remaining children, 2 are girls.)
Since we want all of these events to happen, we MULTIPLY:
3/6 * 2/5 * 3/4 * 2/3 = 1/10.

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 represents one way to get 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 * 1/10 = 3/5.

The correct answer is D.

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 sana.noor » Fri Aug 30, 2013 3:29 am
1) 6c4 = 15
2) selecting 2 girls= 3c2= 3
3) selecting 2 boys = 3c2 =3

3.3/15 = 9/15 = 3/5 D is the answer
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by info2 » Fri Nov 11, 2016 7:33 am
GMATGuruNY wrote:
viskhot wrote:Q:

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
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 to select 2 boys, then 2 girls:
BBGG.
P(B on the 1st pick) = 3/6. (Of the 6 children, 3 are boys.)
P(B on the 2nd pick) = 2/5. (Of the 5 remaining children, 2 are boys.)
P(G on the 3rd pick) = 3/4. (Of the 4 remaining children, 3 are girls.)
P(G on the last pick) = 2/3. (Of the 3 remaining children, 2 are girls.)
Since we want all of these events to happen, we MULTIPLY:
3/6 * 2/5 * 3/4 * 2/3 = 1/10.

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 represents one way to get 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 * 1/10 = 3/5.

The correct answer is D.

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
Hello Mitch

Can you explain which combinations would repeat? 3 boys and 3 girls are all different. I am unable to understand which combination would be identical.

Thanks

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by [email protected] » Fri Nov 11, 2016 9:39 am
Hi All,

A certain number of Quant questions on the GMAT can be solved fairly easily by using "brute force." While the technique isn't 'elegant', it can be done rather quickly in certain situations (and would be preferable to just staring at the question and hoping that an idea comes to you...

Here, we have 3 boys and 3 girls. Let's label them...

Boys: A, B and C
Girls: 1, 2 and 3

We're forming groups of 4 from a total of 6 children. Since we're dealing with GROUPS, the order does NOT matter. Thus, there are...

6!/4!(6-4)! = (6)(5)/(2)(1) = 15 possible groups

We're asked for the probability of an equal number of boys and girls appearing in the group of 4. Logically, we're going to end up with some integer fraction out of 15, so we can eliminate the first 3 answer choices (since you can't reduced down to those denominators). Now, let's just list out the 15 options...

ABC1
ABC2
ABC3
AB12
AB13
AB23
AC12
AC13
AC23
A123

BC12
BC13
BC23
B123

C123

Of the 15 options, how many have exactly 2 boys and 2 girls? Nine.

Final Answer: [spoiler]9/15 - D[/spoiler]

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by info2 » Fri Nov 11, 2016 10:05 am
[email protected] wrote:Hi All,

A certain number of Quant questions on the GMAT can be solved fairly easily by using "brute force." While the technique isn't 'elegant', it can be done rather quickly in certain situations (and would be preferable to just staring at the question and hoping that an idea comes to you...

Here, we have 3 boys and 3 girls. Let's label them...

Boys: A, B and C
Girls: 1, 2 and 3

We're forming groups of 4 from a total of 6 children. Since we're dealing with GROUPS, the order does NOT matter. Thus, there are...

6!/4!(6-4)! = (6)(5)/(2)(1) = 15 possible groups

We're asked for the probability of an equal number of boys and girls appearing in the group of 4. Logically, we're going to end up with some integer fraction out of 15, so we can eliminate the first 3 answer choices (since you can't reduced down to those denominators). Now, let's just list out the 15 options...

ABC1
ABC2
ABC3
AB12
AB13
AB23
AC12
AC13
AC23
A123

BC12
BC13
BC23
B123

C123

Of the 15 options, how many have exactly 2 boys and 2 girls? Nine.

Final Answer: [spoiler]9/15 - D[/spoiler]

GMAT assassins aren't born, they're made,
Rich
Hi Rich

I know combinations approach is the easiest for this question but i want to understand the probability approach in which we dont have to list all the outcomes.

Probability of selecting 2 boys and 2 girls

3/6*2/6*3/4*2/3 * 4!/2!2! =3/5

we are already considering the order when we are selecting 2 boys and 2 girls above. i dont understand the part in bold where we are multiplying by 4! to arrange 4 objects and dividing by 2! twice to account for the duplicates. Which duplicates are there i am confused by this. Can you help understand?

Thanks

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by GMATGuruNY » Fri Nov 11, 2016 10:22 am
info2 wrote:I know combinations approach is the easiest for this question but i want to understand the probability approach in which we dont have to list all the outcomes.

Probability of selecting 2 boys and 2 girls

3/6*2/6*3/4*2/3 * 4!/2!2! =3/5

we are already considering the order when we are selecting 2 boys and 2 girls above. i dont understand the part in bold where we are multiplying by 4! to arrange 4 objects and dividing by 2! twice to account for the duplicates. Which duplicates are there i am confused by this. Can you help understand?

Thanks
P(BBGG) = 3/6 * 2/5 * 3/4 * 2/3 = 1/10.
The fraction in blue represents the probability that the first two selections are boys and the last two selections are girls.

Now we must account for ALL of the ways to select exactly 2 boys and 2 girls.
Here are ALL the ways to select exactly two boys and exactly two girls:
BBGG
BGBG
BGGB
GBBG
GBGB
GGBB.

Total ways = 6.

Since there are 6 ways to select exactly 2 boys and exactly 2 girls, the blue fraction above must be multiplied by 6:
1/10 * 6 = 3/5.

The 6 ways in red constitute the number of ways to arrange the letters BBGG:
4!/2!2! = 6.
I used this approach in my solution above.
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by info2 » Fri Nov 11, 2016 11:16 am
Hi Mitch

I think now i understand the 4!/2!2! part.


3/6. 2/5. 3/4.2/3 here order within the boys has already been accounted for . same thing with the 2 girls group. When we multiply by 4! it will duplicate groups of boys and girls which have already been considered in the 3/6. 2/5. 3/4.2/3 part.

e.g B1B2GiG2 would be repeated when we multiply by 4! so to eliminate those repeated we need to divide by 2! twice.

Is this reasoning correct?


Thanks

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by GMATGuruNY » Fri Nov 11, 2016 12:07 pm
info2 wrote:Hi Mitch

I think now i understand the 4!/2!2! part.


3/6. 2/5. 3/4.2/3 here order within the boys has already been accounted for . same thing with the 2 girls group. When we multiply by 4! it will duplicate groups of boys and girls which have already been considered in the 3/6. 2/5. 3/4.2/3 part.

e.g B1B2GiG2 would be repeated when we multiply by 4! so to eliminate those repeated we need to divide by 2! twice.

Is this reasoning correct?


Thanks
Let's examine all of the ways to select exactly 2 boys and exactly 2 girls.

Case 1: Probability that the first and second selections are boys and the third and fourth selections are girls
P(BBGG) = 3/6 * 2/5 * 3/4 * 2/3 = 1/10.

Case 2: Probability that the first and second selections are boys and the second and fourth selections are girls
P(BGBG) = 3/6 * 3/5 * 2/4 * 2/3 = 1/10.

Case 3: Probability that the first and fourth selections are boys and the second and third selections are girls
P(BGGB) = 3/6 * 3/5 * 2/4 * 2/4 = 1/10.

Case 4: Probability that the second and third selections are boys and the first and fourth selections are girls
P(GBBG) = 3/6 * 3/5 * 2/4 * 2/3 = 1/10.

Case 5: Probability that the second and fourth selections are boys and the first and third selections are girls
P(GBGB) = 3/6 * 3/5 * 2/4 * 2/3 = 1/10.

Case 6: Probability that the third and fourth selections are boys and the first and second selections are girls
P(GGBB) = 3/6 * 2/5 * 3/4 * 2/3 = 1/10.

Every case above represents a different way to select exactly two boys and exactly two girls.
Thus, the probability of selecting exactly 2 boys and exactly 2 girls is equal to the SUM of the fractions above:
1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 = 1/10 * 6 = 3/5.

Notice the following:
The probability in each case is THE SAME (1/10).
The number of different cases -- 6 -- is equal to the number of ways to arrange the letters BBGG:
4!/2!2! = 6.
This is why we multiply the probability of Case 1 (3/6 * 2/5 * 3/4 * 2/3) by the number of ways to arrange the letters BBGG (4!/2!2!):
(3/6 * 2/5 * 3/4 * 2/3) * (4!/2!2!) = 1/10 * 6 = 3/5.
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by Matt@VeritasPrep » Fri Nov 11, 2016 1:58 pm
info2 wrote:I know combinations approach is the easiest for this question but i want to understand the probability approach in which we dont have to list all the outcomes.
Here's another way.

Since we're picking four kids out of a group of six, it's impossible for the group to be all girls or all boys: we must have at least one of each. So our group possibilities are:

* Three boys, one girl
* Three girls, one boy
* Two boys, two girls

With that in mind, we can solve by calculating 1 - P(three boys) - P(three girls).

If we're picking three kids out of our group of 6, we can pick all three boys in exactly one way, and one girl in exactly three ways, for 1 * 3 valid outcomes out of (6 choose 4) = 15 total outcomes.

That gives us P(three boys) = 3/15 = 1/5. P(three girls) is the same = 1/5.

Our answer is thus 1 - 1/5 - 1/5 => 3/5.

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by Scott@TargetTestPrep » Sat Nov 12, 2016 5:37 am
viskhot wrote:Q:

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
We are given that from a group of 3 boys and 3 girls, 4 children are to be randomly selected. We need to determine the probability that an equal numbers of boys and girls will be selected, that is, the probability that two boys and two girls are selected.

We can use combinations to determine the number of favorable outcomes (that 2 boys and 2 girls are selected) and the total number of outcomes (that 4 children are selected from 6 children).

Let's first determine the number of ways we can select 2 boys from 3 boys and 2 girls from 3 girls.

# of ways to select 2 boys from a total of 3 boys: 3C2 = 3

# of ways to select 2 girls from a total of 3 girls: 3C2 = 3

Thus, the number of ways to select 2 girls and 2 boys = 3 x 3 = 9.

Now we can determine the total number of ways to select 4 children from a total of 6 children.

6C4 = (6 x 5 x 4 x 3)/(4 x 3 x 2 x 1) = 3 x 5 = 15

Thus, the probability of selecting an equal number of girls and boys is 9/15 = 3/5.

Answer:D

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