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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pm; Thanked: 11 times; Followed by:8 members

Probability question

by gmattesttaker2 » Mon Jun 18, 2012 9:57 pm

Hello,

I came across the following question that was posted in this forum some time ago:

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

These were the answer choices:

8/28 9/28 10/28 10/18 11/18

The correct answer was given as 9/28.

The way I am trying to solve this is as follows:

( P(E) = Probability for the event to occur
n(E) = Number of ways the event could occur
n(S) = Total number of possible outcomes )

P(E) = P(E1). P(E2). P(E3)
= (n(E1)/n(S1)). (n(E2)/n(S2)). (n(E3)/n(S3))
= (2C1/8C1).(3C2/8C2).(3C2/8C2)

However, I am not sure if this approach is correct since I am not getting 9/28.

I was wondering if you can please explain why I am going wrong here? Thanks for your help.

Best Regards,
Sri

Quote

Source: Beat The GMAT — Problem Solving | Mon Jun 18, 2012 9:57 pm

gmat_and_me: Senior | Next Rank: 100 Posts; Posts: 58; Joined: Sat Mar 05, 2011 9:14 am; Location: Bangalore; Thanked: 20 times; Followed by:5 members; GMAT Score:770

by gmat_and_me » Tue Jun 19, 2012 7:15 am

This approach doesn't seem correct. First of all this is a problem
without replacement. That means that every time you pick a ball,
the denominator for calculation of probability has to decrease by 1.

So, _as per your method_, the probability will have to be

2/8 * 3/7 * 2/6 * 3/5 * 2/4 = 3/280 (in the order R, G, G, B, B).
But that is only one arrangement. The resultant probability is probability
of getting R, G, G, B & B in that order. What about getting R, G, B, G, B ?
Again P is same. But there is one more arrangement to the original.
So there are more such possibilities. In fact, if you fix position of
R as 1, then G, G, B, B can be arranged in (4 * 3 * 2 * 1)/ (2 * 2) = 6
ways. Now R can move from positions 1 to 5 and that means that there are
30 such arrangements. So probability will be 30 * 3/280 = 9/28

Another way to look at the problem:

Total possibilities = 8 * 7 * 6 * 5 * 4

Now if there were only 5 balls, number of ways of selecting them
will be 5 * 4 * 3 * 2 * 1 (i.e R, G, G, B, B). For each of these
possibilities, you have one more duplicate because there are 2
reds. Hence the result now becomes 2 * (5 * 4 * 3 * 2 * 1). We
can select 2 greens in 3c2 ways => 3 more ways. Similarly for
Blue, there are 3 more ways (3c2).

So P = (2 * 3 * 3)(5 * 4 * 3 * 2 * 1)/(8 * 7 * 6 * 5 * 4)

HTH

gmattesttaker2 wrote:Hello,

I came across the following question that was posted in this forum some time ago:

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

These were the answer choices:

8/28 9/28 10/28 10/18 11/18

The correct answer was given as 9/28.

The way I am trying to solve this is as follows:

( P(E) = Probability for the event to occur
n(E) = Number of ways the event could occur
n(S) = Total number of possible outcomes )

P(E) = P(E1). P(E2). P(E3)
= (n(E1)/n(S1)). (n(E2)/n(S2)). (n(E3)/n(S3))
= (2C1/8C1).(3C2/8C2).(3C2/8C2)

However, I am not sure if this approach is correct since I am not getting 9/28.

I was wondering if you can please explain why I am going wrong here? Thanks for your help.

Best Regards,
Sri

Quote

GMAT/MBA Expert

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

by Brent@GMATPrepNow » Tue Jun 19, 2012 7:33 am

gmattesttaker2 wrote: Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

8/28 9/28 10/28 10/18 11/18

Think of the 8 balls as G1, G2, G3, R1, R2, B1, B2, and B3

# of ways to select 5 balls (in any order)
= 8C5
= 56

# of ways to select 1 red, 2 green and 2 blue
- # of ways to select 1 red ball from 2 red balls (R1 and R2) = 2C1 = 2
- # of ways to select 2 green balls from 3 green balls = 3C2 = 3
- # of ways to select 2 blue balls from 3 blue balls = 3C2 = 3
So, total # of ways to select 1 red, 2 green and 2 blue = 2x3x3 = 18

P(select 1 red, 2 green and 2 blue) = [spoiler]18/56 = 9/28[/spoiler]

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Quote

dhonu121: Master | Next Rank: 500 Posts; Posts: 316; Joined: Sun Aug 21, 2011 6:18 am; Thanked: 16 times; Followed by:6 members

by dhonu121 » Tue Jun 19, 2012 9:56 am

Hi Brent,
In your solution,
1.How do you accomodate the fact that the balls are taken out without replacement ?
2.What have you done in your solution above, if int the question, balls were taken out with replacemnet ?

If you've liked my post, let me know by pressing the thanks button.

Quote

GMAT/MBA Expert

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

by Brent@GMATPrepNow » Tue Jun 19, 2012 10:09 am

dhonu121 wrote:Hi Brent,
In your solution,
1.How do you accomodate the fact that the balls are taken out without replacement ?
2.What have you done in your solution above, if int the question, balls were taken out with replacemnet ?

1. In the denominator, I'm looking for the number of ways to select 5 of the 8 balls. This does not allow for replacement. The same applies to the numerator.

2. This would require several steps. If you want to post it as a separate question, I'd be happy to answer it in a new thread. For the curious out there, if we allow for replacement each time, the probability = (30)(1/4)(3/8)(3/8)(3/8)(3/8)

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Quote

dhonu121: Master | Next Rank: 500 Posts; Posts: 316; Joined: Sun Aug 21, 2011 6:18 am; Thanked: 16 times; Followed by:6 members

by dhonu121 » Tue Jun 19, 2012 10:18 am

Brent@GMATPrepNow wrote:
dhonu121 wrote:Hi Brent,
In your solution,
1.How do you accomodate the fact that the balls are taken out without replacement ?
2.What have you done in your solution above, if int the question, balls were taken out with replacemnet ?
1. In the denominator, I'm looking for the number of ways to select 5 of the 8 balls. This does not allow for replacement. The same applies to the numerator.

2. This would require several steps. If you want to post it as a separate question, I'd be happy to answer it in a new thread. For the curious out there, if we allow for replacement each time, the probability = (30)(1/4)(3/8)(3/8)(3/8)(3/8)

Cheers,
Brent

Ok. The catch is that when we use the nCr formula, it does not take into account replacement.
That was new to me.

I used the method used by you to answer my 2nd question to solve the first part.

Thanks for answering.

If you've liked my post, let me know by pressing the thanks button.

Quote

gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pm; Thanked: 11 times; Followed by:8 members

by gmattesttaker2 » Wed Jun 20, 2012 4:51 pm

gmat_and_me wrote:This approach doesn't seem correct. First of all this is a problem
without replacement. That means that every time you pick a ball,
the denominator for calculation of probability has to decrease by 1.

So, _as per your method_, the probability will have to be

2/8 * 3/7 * 2/6 * 3/5 * 2/4 = 3/280 (in the order R, G, G, B, B).
But that is only one arrangement. The resultant probability is probability
of getting R, G, G, B & B in that order. What about getting R, G, B, G, B ?
Again P is same. But there is one more arrangement to the original.
So there are more such possibilities. In fact, if you fix position of
R as 1, then G, G, B, B can be arranged in (4 * 3 * 2 * 1)/ (2 * 2) = 6
ways. Now R can move from positions 1 to 5 and that means that there are
30 such arrangements. So probability will be 30 * 3/280 = 9/28

Another way to look at the problem:

Total possibilities = 8 * 7 * 6 * 5 * 4

Now if there were only 5 balls, number of ways of selecting them
will be 5 * 4 * 3 * 2 * 1 (i.e R, G, G, B, B). For each of these
possibilities, you have one more duplicate because there are 2
reds. Hence the result now becomes 2 * (5 * 4 * 3 * 2 * 1). We
can select 2 greens in 3c2 ways => 3 more ways. Similarly for
Blue, there are 3 more ways (3c2).

So P = (2 * 3 * 3)(5 * 4 * 3 * 2 * 1)/(8 * 7 * 6 * 5 * 4)

HTH

gmattesttaker2 wrote:Hello,

I came across the following question that was posted in this forum some time ago:

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

These were the answer choices:

8/28 9/28 10/28 10/18 11/18

The correct answer was given as 9/28.

The way I am trying to solve this is as follows:

( P(E) = Probability for the event to occur
n(E) = Number of ways the event could occur
n(S) = Total number of possible outcomes )

P(E) = P(E1). P(E2). P(E3)
= (n(E1)/n(S1)). (n(E2)/n(S2)). (n(E3)/n(S3))
= (2C1/8C1).(3C2/8C2).(3C2/8C2)

However, I am not sure if this approach is correct since I am not getting 9/28.

I was wondering if you can please explain why I am going wrong here? Thanks for your help.

Best Regards,
Sri

Hello gmat_and_me,

The explanation for getting 3/280 is clear now. Thanks for the same.

I though had a question about the following:

In fact, if you fix position of
R as 1, then G, G, B, B can be arranged in (4 * 3 * 2 * 1)/ (2 * 2) = 6

I wasn't clear about the 6 arrangements though.

I first listed the Greens and Blues separtely as follows:

G1 G2 B1 B2
G1 G2 B2 B1
G1 B1 G2 B2
G1 B1 B2 G2
G1 B2 G2 B1
G1 B2 B1 G2

G2 G1 B1 B2
G2 G1 B2 B1
G2 B1 G1 B2
G2 B1 B2 G1
G2 B2 G1 B1
G2 B2 B1 G1

B1 G1 G2 B2
B1 G1 B2 G2
B1 G2 G1 B2
B1 G2 B2 G1
B1 B2 G1 G2
B1 B2 G2 G1

B2 G1 G2 B1
B2 G1 B1 G2
B2 G2 G1 B1
B2 G2 B1 G1
B2 B1 G1 G2
B2 B1 G2 G1

Eliminating duplicates I got the following:

G G B B
G B G B
G B B G

B B G G
B G B G
B G G B

But I am kind of stuck at this point. I was wondering if you can please help here? Thanks again for your valuable time and help.

Best Regards,
Sri

Quote

GMAT/MBA Expert

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

by Brent@GMATPrepNow » Thu Jun 21, 2012 5:55 am

gmattesttaker2 wrote: Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

A) 8/28
B) 9/28
C) 10/28
D) 10/18
E) 11/18

We can also solve this question using a combination of probability rules and counting methods:

Let's examine one possible outcome: RGGBB (the first ball is red, the second ball is green, the third ball is green, etc.)

What is the probability of this particular outcome?

Well, P(RGGBB) = P(red on 1st AND green on 2nd AND green on 3rd AND blue on 4th AND blue on 5th)
= P(red on 1st) x P(green on 2nd) x P(green on 3rd) x P(blue on 4th) x P(blue on 5th)
= (2/8)(3/7)(2/6)(3/5)(2/4)
= 3/280

Of course this is just the probability of one possible outcome. We need to consider others. For example, the outcome BGRGB also meets the criteria.

So, P(BGRGB) = P(blue on 1st AND green on 2nd AND red on 3rd AND green on 4th AND blue on 5th)
= P(blue on 1st) x P(green on 2nd) x P(red on 3rd) x P(green on 4th) x P(blue on 5th)
= (3/8)(3/7)(2/6)(2/5)(2/4)
= 3/280

Important: The probability of each acceptable outcome will equal 3/280. So the question now becomes, "In how many ways can we draw 1 red, 2 green and 2 blue balls?" In other words, in how many different ways can we arrange the letters in RGGBB?

This is what is known as a MISSISSIPI question, where we are arranging objects where some of the objects are identical.

Here we have 5 objects (letters) and we have 2 identical letters, and another 2 identical letters. So, we can arrange these 5 objects is 5!/2!2! ways. This is equal to 30

So there are 30 different ways to draw 1 red, 2 green and 2 blue balls, and the probability that each one occurs = 3/28

So, P(drawing 1 red, 2 green and 2 blue) = [spoiler](30)(3/28) = 9/28 = B[/spoiler]

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Quote

gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pm; Thanked: 11 times; Followed by:8 members

by gmattesttaker2 » Thu Jun 21, 2012 1:13 pm

Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

A) 8/28
B) 9/28
C) 10/28
D) 10/18
E) 11/18

We can also solve this question using a combination of probability rules and counting methods:

Let's examine one possible outcome: RGGBB (the first ball is red, the second ball is green, the third ball is green, etc.)

What is the probability of this particular outcome?

Well, P(RGGBB) = P(red on 1st AND green on 2nd AND green on 3rd AND blue on 4th AND blue on 5th)
= P(red on 1st) x P(green on 2nd) x P(green on 3rd) x P(blue on 4th) x P(blue on 5th)
= (2/8)(3/7)(2/6)(3/5)(2/4)
= 3/280

Of course this is just the probability of one possible outcome. We need to consider others. For example, the outcome BGRGB also meets the criteria.

So, P(BGRGB) = P(blue on 1st AND green on 2nd AND red on 3rd AND green on 4th AND blue on 5th)
= P(blue on 1st) x P(green on 2nd) x P(red on 3rd) x P(green on 4th) x P(blue on 5th)
= (3/8)(3/7)(2/6)(2/5)(2/4)
= 3/280

Important: The probability of each acceptable outcome will equal 3/280. So the question now becomes, "In how many ways can we draw 1 red, 2 green and 2 blue balls?" In other words, in how many different ways can we arrange the letters in RGGBB?

This is what is known as a MISSISSIPI question, where we are arranging objects where some of the objects are identical.

Here we have 5 objects (letters) and we have 2 identical letters, and another 2 identical letters. So, we can arrange these 5 objects is 5!/2!2! ways. This is equal to 30

So there are 30 different ways to draw 1 red, 2 green and 2 blue balls, and the probability that each one occurs = 3/28

So, P(drawing 1 red, 2 green and 2 blue) = [spoiler](30)(3/28) = 9/28 = B[/spoiler]

Cheers,
Brent

Hello Brent,

Thank you very much for your detailed explanation. It is a lot clearer now. Is this approach called the Counting/Permutations method? Also, the other approach that you have described in your earlier post here, is it the Combinations method :

Think of the 8 balls as G1, G2, G3, R1, R2, B1, B2, and B3

# of ways to select 5 balls (in any order)
= 8C5
= 56

# of ways to select 1 red, 2 green and 2 blue
- # of ways to select 1 red ball from 2 red balls (R1 and R2) = 2C1 = 2
- # of ways to select 2 green balls from 3 green balls = 3C2 = 3
- # of ways to select 2 blue balls from 3 blue balls = 3C2 = 3
So, total # of ways to select 1 red, 2 green and 2 blue = 2x3x3 = 18

P(select 1 red, 2 green and 2 blue) = 18/56 = 9/28

Now in this example both the above methods work great. However, for something like the Goodbuddies problem that I saw in this forum a few times:

https://www.beatthegmat.com/probability-t22021.html

will the Combination method also work. I was able to solve it via the counting method since I personally felt it was a bit easier. Thanks again for all your valuable time and help.

Best Regards,
Sri

Quote

GMAT/MBA Expert

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

by Brent@GMATPrepNow » Thu Jun 21, 2012 4:52 pm

The first method I used was primarily an application of the Fundamental Counting Principle (FCP). . . along with some combinations.

Aside: People also refer to the application of the FCP as the "slot method"

The second method I used would be best described as using the rules of probability.

If you're interested, you can watch a free video about the FCP here: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Quote

gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pm; Thanked: 11 times; Followed by:8 members

by gmattesttaker2 » Thu Jun 21, 2012 7:46 pm

Brent@GMATPrepNow wrote:The first method I used was primarily an application of the Fundamental Counting Principle (FCP). . . along with some combinations.

Aside: People also refer to the application of the FCP as the "slot method"

The second method I used would be best described as using the rules of probability.

If you're interested, you can watch a free video about the FCP here: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

Hello Brent,

Thank you very much for the explanation and also for the tutorial link. Thanks again.

Best Regards,
Sri

Quote

dhonu121: Master | Next Rank: 500 Posts; Posts: 316; Joined: Sun Aug 21, 2011 6:18 am; Thanked: 16 times; Followed by:6 members

by dhonu121 » Thu Jun 21, 2012 10:16 pm

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

These were the answer choices:

8/28 9/28 10/28 10/18 11/18

Please be reminded that nCr gives the number of selections of r objects from n objects without replacement.
Since the case at hand considers without replacement we can use this formula.

So denominator = 8C5 = 56
Numerator = (2C1*3C2*3C2) = 18
Thus prob = 18/56 = 9/28

If you've liked my post, let me know by pressing the thanks button.

Quote

parveen110: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Fri Jan 17, 2014 7:34 am; Thanked: 7 times

by parveen110 » Sat May 31, 2014 1:13 am

Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

8/28 9/28 10/28 10/18 11/18
Think of the 8 balls as G1, G2, G3, R1, R2, B1, B2, and B3

# of ways to select 5 balls (in any order)
= 8C5
= 56

# of ways to select 1 red, 2 green and 2 blue
- # of ways to select 1 red ball from 2 red balls (R1 and R2) = 2C1 = 2
- # of ways to select 2 green balls from 3 green balls = 3C2 = 3
- # of ways to select 2 blue balls from 3 blue balls = 3C2 = 3
So, total # of ways to select 1 red, 2 green and 2 blue = 2x3x3 = 18

P(select 1 red, 2 green and 2 blue) = [spoiler]18/56 = 9/28[/spoiler]

Cheers,
Brent

Hi Brent!!

I understand from your posts that using FCP is the most effective way to solve probablity and P&C questions on GMAT. However, I find it hard to feel intuitive about FCP.

For e.g. In the approach you've used above,# of ways to select 2 blue balls from 3 blue balls = 3C2

But if 3 blue balls (say, b1,b2 and b3) are identical then, it shouldn't matter whether i select b1, b2 or b2,b3 or b3,b1. I mean, Should i not have just one option to select 2 blue balls from 3 identical blue balls?

Your second approach was how i solved it for the first time. But it's a bit longer approach.