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Probability of picking flowers

by topspin360 » Wed Jul 04, 2012 1:31 pm
Can someone please explain why the answer to the following question is 13/18 AND MORE IMPORTANTLY why it isn't 11/21 [1 - (2C2 + 3C2 + 4C2)/7C2].

A florist has 2 azaleas, 3 buttercups, and 4 petunias. If she puts two flowers in a bouquet, what's the probability that the two flower are not the same type?

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by GMATGuruNY » Wed Jul 04, 2012 2:06 pm
topspin360 wrote:Can someone please explain why the answer to the following question is 13/18 AND MORE IMPORTANTLY why it isn't 11/21 [1 - (2C2 + 3C2 + 4C2)/7C2].

A florist has 2 azaleas, 3 buttercups, and 4 petunias. If she puts two flowers in a bouquet, what's the probability that the two flower are not the same type?

Thanks.
P(2 different flowers) = 1 - P(2 of the same flower).

P(2 azaleas):
P(1st flower is an azalea) = 2/9. (Of the 9 flowers, 2 are azaleas.)
P(2nd flower is an azalea) = 1/8. (Of the 8 remaining flowers, 1 is an azalea.)
Since we want both events to happen, we multiply the fractions:
2/9 * 1/8 = 1/36.

P(2 buttercups):
P(1st flower is a buttercup) = 3/9. (Of the 9 flowers, 3 are buttercups.)
P(2nd flower is a buttercup) = 2/8. (Of the 8 remaining flowers, 2 are buttercups.)
Since we want both events to happen, we multiply the fractions:
3/9 * 2/8 = 1/12.

P(2 petunias):
P(1st flower is a petunia) = 4/9. (Of the 9 flowers, 4 are petunias.)
P(2nd flower is a petunia) = 3/8. (Of the 8 remaining flowers, 3 are petunias.)
Since we want both events to happen, we multiply the fractions:
4/9 * 3/8 = 1/6.

Since any of the above outcomes would yield 2 of the same flower, we add the fractions:
P(2 of the same flower) = 1/36 + 1/12 + 1/6 = 10/36 = 5/18.

Thus, P(2 different flowers) = 1 - 5/18 = 13/18.
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by eagleeye » Wed Jul 04, 2012 2:33 pm
topspin360 wrote:Can someone please explain why the answer to the following question is 13/18 AND MORE IMPORTANTLY why it isn't 11/21 [1 - (2C2 + 3C2 + 4C2)/7C2].

A florist has 2 azaleas, 3 buttercups, and 4 petunias. If she puts two flowers in a bouquet, what's the probability that the two flower are not the same type?

Thanks.
We should make that payment arrangement right about now. jk. ;)
Topspin360, you made a couple of algebraic errors, the concept behind your calculation was correct.


What you should have calculated was: probability = 1- (2C2+3C2+4C2)/(9C2) = 1 - (1+3+6)/36 = 1- 10/36= 1- 5/18 = 13/18. So your mind was in the right place. I am guessing you made the following errors:
1) Added 2+3+4 to be 7 instead of 9.
2) Took 2C2 = 2 instead of 1.

Careless errors. But the concept was fine.
Here's another way of calculating it using fundamental principle of counting. However, the way you did it is pretty good.

Add the probabilities of the 3 cases when the first flower is A, B, or P.

Total required probability. = (2/9)*(7/8) + (3/9)*(6/8) + (4/9)*(5/8) = (14+18+20)/72 = 52/72 = 13/18.

Let me know if this helps :)
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by theCEO » Wed Jul 04, 2012 3:03 pm
topspin360 wrote:Can someone please explain why the answer to the following question is 13/18 AND MORE IMPORTANTLY why it isn't 11/21 [1 - (2C2 + 3C2 + 4C2)/7C2].

A florist has 2 azaleas, 3 buttercups, and 4 petunias. If she puts two flowers in a bouquet, what's the probability that the two flower are not the same type?

Thanks.

Alternative approach:

Find all the total possibilities of picking 2 flowers: (9 * 8)/2 = 36
There are 36 arrangements that can be made.

Ways to arrange azaleas:
Since there are 2 azaleas, there is only 1 way they can be arranged (2x1/2 or AB)

Ways to arrange buttercups:
Since there are 3 buttercups, there are 3 ways we can arrange them (AB, AC, BC)

Ways to arrange petunias
Since there are 4 petunias,there are 6 was to arrange them (AB,AC,AD,BC,BD,CD)

Total possibilities = same flowers arrangements + different flowers arrangements
36 = (1+3+6) + different arrangements
different arrangements = 36 - 10 = 26

Probability of different arrangement = total different arrangement / total possibilty
Probability of different arrangement = 26/36 = 13/18
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by topspin360 » Wed Jul 04, 2012 4:23 pm
Thanks all for the help.

eagleeye, I'm open to the idea haha. PM me if you're interested in providing tutor services at a decent rate.
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by GMATGuruNY » Wed Jul 04, 2012 6:09 pm
topspin360 wrote:Can someone please explain why the answer to the following question is 13/18 AND MORE IMPORTANTLY why it isn't 11/21 [1 - (2C2 + 3C2 + 4C2)/7C2].

A florist has 2 azaleas, 3 buttercups, and 4 petunias. If she puts two flowers in a bouquet, what's the probability that the two flower are not the same type?

Thanks.
One more approach:

Case 1: Azalea on the first pick, different flower on the second pick
P(1st flower is an azalea) = 2/9. (Of the 9 flowers, 2 are azaleas.)
P(2nd flower is not an azalea) = 7/8. (Of the 8 remaining flowers, 7 are not azaleas.)
Since we want both events to happen, we multiply the fractions:
2/9 * 7/8 = 7/36.

Case 2: Buttercup on the first pick, different flower on the second pick
P(1st flower is a buttercup) = 3/9. (Of the 9 flowers, 3 are buttercups.)
P(2nd flower is not a buttercup) = 6/8. (Of the 8 remaining flowers, 6 are not buttercups.)
Since we want both events to happen, we multiply the fractions:
3/9 * 6/8 = 9/36.

Case 3: Petunia on the first pick, different flower on the second pick
P(1st flower is a petunia) = 4/9. (Of the 9 flowers, 4 are petunias.)
P(2nd flower is not a petunia) = 5/8. (Of the 8 remaining flowers, 5 are not petunias.)
Since we want both events to happen, we multiply the fractions:
4/9 * 5/8 = 10/36.

Since any of the above outcomes would yield 2 different types of flowers, we add the fractions:
7/36 + 9/36 + 10/36 = 26/36 = 13/18.
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by talaangoshtari » Tue Jun 23, 2015 9:37 am
Would you please explain why this approach is wrong?
the probability that the two flower are not the same type:
P[(a,b) or (a,p) or (b,p)]
=(2/9)(3/9)+(2/9)(4/9)+(3/9)(4/9)=26/81
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by Brent@GMATPrepNow » Tue Jun 23, 2015 9:50 am
A florist has 2 azaleas, 3 buttercups and 4 petunias. She puts two flowers together at a random in a bouquet. However, the customer calls and tells she does not want the two of d same flower. What is the probability that the florist doesnt have to change the bouquet?
First, we can REWRITE the question as "What is the probability that the two flowers are DIFFERENT colors?"
This is a good candidate to use the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)

When applied to this questions, we get: P(DIFFERENT 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

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

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(DIFFERENT colors) = 1 - P(SAME color)
= 1 - 5/18
= [spoiler]13/18[/spoiler]

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by Brent@GMATPrepNow » Tue Jun 23, 2015 9:54 am
talaangoshtari wrote:Would you please explain why this approach is wrong?
the probability that the two flower are not the same type:
P[(a,b) or (a,p) or (b,p)]
=(2/9)(3/9)+(2/9)(4/9)+(3/9)(4/9)=26/81
To begin, once you choose 1 flower, there are only 8 flowers remaining.
So, for example, P(a then b) = (2/9)(3/8) [you have (2/9)(3/9)]

Also, what about P(b then a)?
And P(a then p)?

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by MartyMurray » Tue Jun 23, 2015 10:06 am
talaangoshtari wrote:Would you please explain why this approach is wrong?
the probability that the two flower are not the same type:
P[(a,b) or (a,p) or (b,p)]
=(2/9)(3/9)+(2/9)(4/9)+(3/9)(4/9)=26/81
When you do probabilities of selecting things, even if they are all SUPPOSEDLY selected at the SAME TIME, you need to do the math as if one were selected after another.

So to further explain what Brent said about there being only 8 flowers remaining after you select one, your calculations would need to be based on first choosing 1 from 9 and then choosing 1 from 8, as in P[(a,b) or (a,p) or (b,p)] = (2/9)(3/8)+(2/9)(4/8)+(3/9)(4/8).

Then, to get it right you would also need to account for other combinations...
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by talaangoshtari » Tue Jun 23, 2015 10:23 am
Hi Brent,

why should we count a then b, and b then a, twice? choosing ab or ba are both lead to the same group
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by Brent@GMATPrepNow » Tue Jun 23, 2015 10:34 am
talaangoshtari wrote:Hi Brent,

why should we count a then b, and b then a, twice? choosing ab or ba are both lead to the same group
Not so.
Consider this analogy: A couple has 2 children.
There are 4 possible outcomes:
- Boy then Boy
- Girl then Boy
- Boy then Girl
- Girl then Girl

So, P(1 boy and 1 girl) = 2/4 = 1/2
In other words, P(1 boy and 1 girl) = P(boy then girl) + P(girl then boy)
= (1/2)(1/2) + (1/2)(1/2)
= 1/4 + 1/4
= 1/2

If we apply your logic, P(1 boy and 1 girl) = (1/2)(1/2) = 1/4, which is incorrect.

If I were you, I'd try listing all of the possible outcomes and see what you get.

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by Matt@VeritasPrep » Tue Jun 23, 2015 6:00 pm
Order in probability is counterintuitive enough that it takes a lot of getting used to: it's definitely a good idea to try a few more of these problems until you get a feel for when AB and BA are treated as distinct scenarios. Anything involving kids or dice is a good place to start.
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by talaangoshtari » Wed Jun 24, 2015 3:48 am
Are the flowers in one group distinct? And because they are distinct the order matters? I think in a family there is a difference between the order of the children, but in picking the flowers I still don't understand why the order matters.
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by GMATGuruNY » Wed Jun 24, 2015 5:32 am
talaangoshtari wrote:Are the flowers in one group distinct? And because they are distinct the order matters? I think in a family there is a difference between the order of the children, but in picking the flowers I still don't understand why the order matters.
When we choose from a SINGLE SOURCE without replacement, the probability of a favorable outcome is the same whether the selections are made ALL AT ONCE or ONE AT A TIME.
To account for all of the possible cases, we should proceed under the assumption that the selections are made one at a time.

Here, we are choosing from a single source of flowers.
Thus, we should proceed under the assumption that the 3 flowers are selected ONE AT A TIME.
If the desired outcome is the selection of exactly 2 azaleas, there are 3 ways to achieve a favorable outcome:
AAN: 1st flower is azalea, 2nd flower is azalea, 3rd flower is non-azalea.
ANA: 1st flower is azalea, 2nd flower is non-azalea, 3rd flower is azalea.
NAA: 1st flower is non-azalea, 2nd flower is azalea, 3rd flower is azalea.
Thus:
P(exactly 2 azaleas) = PIAAN) + P(ANA) + P(NAA).
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