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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Probability Question

by e12star » Fri Jan 13, 2012 10:57 am
Question from Word Translations (Manhattan GMAT), chpt 5:
A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts two flowers together at random in a bouquet. However, the customer calls and says that she does not want two of the same flower. What is the probability that the florist does not have to change the bouquet?

I understand how to find the number of different 2-flower bouquets (by using the anagram model). However, is there a mathematical way to find the number of different bouquets in which both flowers are the same (without actually writing them all out)?

Thanks!
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by Mike@Magoosh » Fri Jan 13, 2012 12:14 pm
Hi, there. I'm happy to help with this. :-)

It sounds like you are already comfortable finding the total number of pairs from a set of nine --- 9C2 =36. Of those 36 pairs, how many would have two of the same flower?

Well, we have 2 azaleas, 3 buttercups, and 4 petunias. Here we have three different questions to answer:
a) How many different ways are there to pick 2 azaleas from 2? Just 1
b) How many different ways are there to pick 2 buttercups from 3? 3C2 = 3
c) How many different ways are there to pick 2 petunias from 4? 4C2 = 6

So, there are 1 + 3 + 6 = 10 different pairs that consist of two of the same flower. Probability of two of the same 10/36 = 5/18.

Does that make sense? Please let me know if you have any questions on what I've said here.

Mike :)
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by Brent@GMATPrepNow » Sat Jan 14, 2012 8:53 am
e12star wrote:Question from Word Translations (Manhattan GMAT), chpt 5:
A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts two flowers together at random in a bouquet. However, the customer calls and says that she does not want two of the same flower. What is the probability that the florist does not have to change the bouquet?
I thought I'd mention that we can also solve this question using the rules of probability.

First, we can rewrite the question as "What is the probability that the two flowers are different colors?"

Well, P(diff colors) = 1 - P(same color)

Aside: let A = azalea, let B = buttercup, let P = petunia

P(same color) = P(both A's OR both B's OR both P's)
= P(both A's) + P(both B's) + P(both P's)

Now let's examine each probability:
P(both A's): we need the 1st flower to be an azalea and the 2nd flower to be an azalea
So, P(both A's) = (2/9)(1/8) = 2/72

Similarly: P(both B's) = (3/9)(2/8) = 6/72

And: P(both P's) = (4/9)(3/8) = 12/72

So, P(same color) = (2/72) + (6/72) + (12/72) = 20/72 = 5/18

Now back to the beginning:
P(diff colors) = 1 - P(same color)
= 1 - 5/18
= 13/18

Cheers,
Brent
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by MakeUrTimeCount » Sat Jan 14, 2012 9:36 am
Another Approach:
We can have different flowers only when we have one of the following 3 combinations:
1) azaleas, buttercups
2) azaleas, buttercups
3) buttercups, petunias.

For option1), number of ways : 2C1.3C1/9C2 = 6/36
For option2), number of ways : 2C1.4C1/9C2 = 8/36
For option3), number of ways : 3C1.4C1/9C2 = 12/36
Addiing them all: 26/36 = 13/18
:)
Last edited by MakeUrTimeCount on Sat Jan 14, 2012 11:00 am, edited 1 time in total.
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by ronnie1985 » Sat Jan 14, 2012 10:32 am
I think there is a serious flaw in the reasoning as now the bouquet is already having 2 flowers of the same color. The florist has to take one of the flower, the color of which is not known and replace it with another flower of some other color. This calls for Baye's Theorem.
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by Brent@GMATPrepNow » Sat Jan 14, 2012 11:03 am
ronnie1985 wrote:I think there is a serious flaw in the reasoning as now the bouquet is already having 2 flowers of the same color. The florist has to take one of the flower, the color of which is not known and replace it with another flower of some other color. This calls for Baye's Theorem.
We can use Baye's Theorem here. We'll choose the 1st flower and then choose a diff 2nd flower.

Now, when it comes to choosing the first flower there are 3 cases:

case 1: choose azalea first, choose different flower second
The probability of choosing an azalea first is 2/9
Once we have selected an azalea first, there are 8 flowers remaining (1 A, 3 B's and 4 P's), and we must choose a different flower second.
So, the probability of choosing a different flower second is 7/8
So, P(azalea first and different flower second) = (2/9)(7/8) = 14/72

case 2: choose buttercup first, choose different flower second
The probability of choosing an buttercup first is 3/9
Once we have selected an buttercup first, there are 8 flowers remaining (2 A's, 2 B's and 4 P's), and we must choose a different flower second.
So, the probability of choosing a different flower second is 6/8
So, P(buttercup first and different flower second) = (3/9)(6/8) = 18/72

case 3: choose petunia first, choose different flower second
The probability of choosing an petunia first is 4/9
Once we have selected an petunia first, there are 8 flowers remaining (2 A's, 3 B's and 3 P's), and we must choose a different flower second.
So, the probability of choosing a different flower second is 5/8
So, P(petunia first and different flower second) = (4/9)(5/8) = 20/72


P(2 diff colors) = (14/72) + (18/72) + (20/72)
= 52/72
= 13/18

Cheers,
Brent
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by ArunangsuSahu » Wed Jan 18, 2012 11:01 pm
Easily done by

P(different) = 1-P(same)

P(same)=(2C2+3C2+4C2)/9c2=5/18

P(different)=1-5/18
=13/18
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by Cumulonimbus » Fri Jan 11, 2013 2:46 am
Hi Brent,

I tried it as below:

P(A and B) = (2/9)*(3/8) = 6/72
P(B and P) = (3/9)*(4/8) = 12/72
P(P and A) = (4/9)*(2/8) = 8/72

Probability of choosing different flowers = P(A and B) + P(B and P) + P(P and A)
= (6+12+8)/72 = 26/72 = 13/36.
Why is this method not giving correct answer?
How is this different from Baye's theorem?

Also, these flower types can come in any order - AB, BP, PA, BA, PB, AP - so total 6 cases,
Are all these cases different, is AB really different from BA?

Since correct answer is 13/18, do i need to multiply here by 2 because AB and BA are different?

My confusion arises in deciding when order matters and when it does not.

Another similar question:

"A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?"

Can you please help with this. Are these two questions any different?

KR,
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by Brent@GMATPrepNow » Fri Jan 11, 2013 7:36 am
Cumulonimbus wrote:Hi Brent,

I tried it as below:

P(A and B) = (2/9)*(3/8) = 6/72
P(B and P) = (3/9)*(4/8) = 12/72
P(P and A) = (4/9)*(2/8) = 8/72

Probability of choosing different flowers = P(A and B) + P(B and P) + P(P and A)
= (6+12+8)/72 = 26/72 = 13/36.
Why is this method not giving correct answer?
How is this different from Baye's theorem?

Also, these flower types can come in any order - AB, BP, PA, BA, PB, AP - so total 6 cases,
Are all these cases different, is AB really different from BA?

Since correct answer is 13/18, do i need to multiply here by 2 because AB and BA are different?

My confusion arises in deciding when order matters and when it does not.

Another similar question:

"A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?"

Can you please help with this. Are these two questions any different?

KR,
Whenever I'm trying to find a certain probability, it helps to stop and ask, " What exactly must occur in order for this event to happen?"

So, in your solution you have P(A and B).
What needs to occur here?
This event can occur in two different ways:
1) select A 1st then select B 2nd
2) select B 1st then select A 2nd
Your solution doesn't account for both scenarios.

Fortunately, P(AB) and P(BA) are equal, so once you have one probability, you can just double it in order to find the probability that either scenario occurs.

---------------------------------------------
Your question:
"A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?"

Once again, we'll ask, " What exactly must occur in order for this event to happen?"
We can get one gumball of each color in six different ways:
- BGR
- BRG
- RGB
- RBG
- GBR
- GRB
Notice that 3! = 6. That is, we can arrange the 3 letters (R, G and B) in 3! different ways.

So, P(1 gumball of each color) = P(BGR) + P(BRG) + P(GRB) + P(GBR) + P(RGB) + P(RBG)
Once again, each of the 6 probabilities are equal, so once you find 1 probability, you can just multiply it by 6 (aka 3!)

I hope that helps.

Cheers,
Brent
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by Cumulonimbus » Sat Jan 12, 2013 11:06 am
Hi Brent,
Thanks.
However I still have one point unclear.
How to decide when order matters and it does not.
For this case - as per my understanding - either first choosing an A and then B or first choosing a B and then A, ultimately results in one type of arrangement.
I can understand that if I arrange digits, or alphabets to form some code (say), then this arrangement matter's. That is 54321 is a different code than 14325, but and Azalea and Buttercups make one arrangement.

Below examples are from Manhattan guide.
Example 1:
An office manager must choose five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition is allowed. How many different lock codes are possible?
Manhattan guide way - number of decisions to be made:
5*8*7*6*4=6720.

My way: I tried via Combinatorics,

Number of ways in which 2 odd digits can be chosen from 5 = 5C2 = 10 (this assumes all five are identical, the two chosen ones as well as the three not-chosen ones).
However since the two digits chosen are not identical, I need to multiply it by 2!, number of ways the chosen two digits can be arranged, so
Number of ways in which 2 odd digits can be chosen from 5 = 5C2 = 10*2 = 20

Similarly, for the rest 3 digits, number of ways to choose 3 digits out of 8 = 8C3, and here I need to multiply it by 3! to give = 56*3! = 336,

Thus total no of codes/combination = 20*336 = 6670.

Here order matters.

Example 2:
The I Eta Pi fraternity must choose a delegation of three senior members and two junior members for an annual fraternity conference. If I Eta Pi has 6 senior members and 5 junior members, how many different delegations are possible?

Here answer is simply = 6C3 = 20, because order does not matter here.

Please guide.

KR,
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by Brent@GMATPrepNow » Sun Jan 13, 2013 10:13 am
Cumulonimbus wrote:Hi Brent,

Thanks.
However I still have one point unclear.
How to decide when order matters and it does not.

For this case - as per my understanding - either first choosing an A and then B or first choosing a B and then A, ultimately results in one type of arrangement.
"Does order matter" is a question that applies more to counting questions than it does probability questions.

To determine how to set up a probability question, always begin with a probability featuring words and slowly transition to symbols.

P(selecting 1 azalea and 1 buttercup) = P(selecting an azalea 1st and a buttercup 2nd OR selecting a buttercup 1st and an azalea 2nd )
= P(selecting an azalea 1st and a buttercup 2nd) + P(selecting a buttercup 1st and an azalea 2nd )

You mentioned that first choosing an A and then B results in the same outcome as first choosing a B and then A. As such, we must consider both of them.
In your solution, you only addressed 1 scenario.

Cheers,
Brent
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by Brent@GMATPrepNow » Sun Jan 13, 2013 10:15 am
Cumulonimbus wrote:
Below examples are from Manhattan guide.
Example 1:
An office manager must choose five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition is allowed. How many different lock codes are possible?
Manhattan guide way - number of decisions to be made:
5*8*7*6*4=6720.

My way: I tried via Combinatorics,

Number of ways in which 2 odd digits can be chosen from 5 = 5C2 = 10 (this assumes all five are identical, the two chosen ones as well as the three not-chosen ones).
However since the two digits chosen are not identical, I need to multiply it by 2!, number of ways the chosen two digits can be arranged, so
Number of ways in which 2 odd digits can be chosen from 5 = 5C2 = 10*2 = 20

Similarly, for the rest 3 digits, number of ways to choose 3 digits out of 8 = 8C3, and here I need to multiply it by 3! to give = 56*3! = 336,

Thus total no of codes/combination = 20*336 = 6670.

Here order matters.

Example 2:
The I Eta Pi fraternity must choose a delegation of three senior members and two junior members for an annual fraternity conference. If I Eta Pi has 6 senior members and 5 junior members, how many different delegations are possible?

Here answer is simply = 6C3 = 20, because order does not matter here.

Please guide.

KR,
It might be easier/better to post these as separate posts so that we don't have 1 thread discussing 3 different questions.

Cheers,
Brent
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Probability

by Deepthi Subbu » Fri May 31, 2013 11:07 pm
My soln :

2/9*3/8 + 2/9*4/8 + 3/9*4/8 => 13/36 . However the answer is 13/18. Where am I going wrong?
Since order doesnot matter here, as azalea and then followed by buttercup is the same otherwise.And in addition to this if there are 3 ways to choose the placement, Il be in more trouble.
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by Brent@GMATPrepNow » Sat Jun 01, 2013 6:13 am
Deepthi Subbu wrote:My soln :

2/9*3/8 + 2/9*4/8 + 3/9*4/8 => 13/36 . However the answer is 13/18. Where am I going wrong?
Since order does not matter here, as azalea and then followed by buttercup is the same otherwise.And in addition to this if there are 3 ways to choose the placement, Il be in more trouble.
The way you have set up your solution, the order does, indeed, matter.

For example, what does 2/9*3/8 stand for?
Presumably it represents the probability of getting 1 azalea and 1 buttercup.
The important thing to recognize is that there are two different ways to get 1 azalea and 1 buttercup.
One way: select an azalea first and a buttercup second.
Another way: select a buttercup first and an azalea second.

Your probability calculation of 2/9*3/8 is for the first way (selecting an azalea first and a buttercup second.)
You also need to consider the second way (selecting a buttercup first and an azalea second.)


Here's another way to look at it.
Let's say there are 5 black chips and 1 red chip.
We are selecting 6 (yes 6) chips, and we want to find the probability of selecting a red chip.
Of course, since we're selecting all 6 chips, the probability = 1.
However, if we say that order does not matter (as you have done), we may consider only one of the ways to select the 6 coins. Let's say we calculate the probability of selecting the red chip first. We get:
P(select red chip) = (1/6)(5/5)(4/4)(3/3)(2/2)(1/1) = 1/6
This is wrong, because we have not consider the probability of selecting the red chip second (1/6) and the probability of selecting the red chip third (1/6) and the probability of selecting the red chip fourth (1/6) etc.

I hope that helps.

Cheers,
Brent
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