Probability MGMAT Guide 5

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arpshriv: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Sat Feb 19, 2011 12:23 am

Probability MGMAT Guide 5

by arpshriv » Mon Sep 01, 2014 10:27 pm

I am unable to understand the explanation provided in the book can someone please help me with the solution.

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?

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 am; Location: Chennai, India; Thanked: 83 times; Followed by:5 members

by Uva@90 » Mon Sep 01, 2014 11:23 pm

arpshriv wrote:I am unable to understand the explanation provided in the book can someone please help me with the solution.

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?

Hi

We need to find the probability of bouquet with no two same flower.

So first find the probability of two same flower in boquest and subtract it from 1.

Total number of flowers is 2+3+4 = 9

Probability of two same flower is 2/9*1/8 + 3/9*2/8 +4/9*3/8 = 2/72 +6/72+12/72 = 20/72

Now,
[spoiler]1-20/72[/spoiler] is the answer

Regards,
Uva

Known is a drop Unknown is an Ocean

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

by GMATGuruNY » Tue Sep 02, 2014 2:43 am

arpshriv wrote:I am unable to understand the explanation provided in the book can someone please help me with the solution.

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?

Approach 1:

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 = [spoiler]13/18[/spoiler].

Approach 2:

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 = [spoiler]13/18[/spoiler].

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by Brent@GMATPrepNow » Tue Sep 02, 2014 9:54 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]

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMAT/MBA Expert

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by Brent@GMATPrepNow » Tue Sep 02, 2014 9:57 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?

Here's an approach that does not use the complement.

To calculate P(2 diff colors), we need to consider 3 cases.
That is, P(2 diff colors) = P(azalea 1st then a different flower OR buttercup 1st then a different flower OR petunias 1st then a different flower)
= P(azalea 1st then a different flower) + P(buttercup 1st then a different flower) + P(petunia 1st then a different flower)

Let's examine each probability separately.

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
= [spoiler]13/18[/spoiler]

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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by [email protected] » Tue Sep 02, 2014 11:51 am

Hi arpshriv,

When dealing with these types of questions, it's important to include the answer choices. Sometimes the "math" behind a complex prompt can be avoided (partially or completely) by using the answer choices to your advantage.

As you may have noticed from the various explanations provided, there is more than one way to approach solving this problem. Most GMAT questions are written in that fashion, so that a flexible thinker won't get "stuck" too often. Keep your mind open to alternative approaches; you'd be amazed how easy many GMAT questions can be to solve, IF you know more than one way to approach the prompt.

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

Contact Rich at [email protected]

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pm; Location: New Delhi, India; Thanked: 205 times; Followed by:24 members

by GMATinsight » Tue Sep 02, 2014 8:12 pm

arpshriv wrote:I am unable to understand the explanation provided in the book can someone please help me with the solution.

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?

Fav. Prob = 1 - (prob of picking 2 of same type)

Fav Prob. = 1- [(2C2 + 3C2 + 4C2) / 9C2] = 13/18

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