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Probability Problem Sets

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by theachiever » Wed Oct 31, 2012 11:44 pm
A florist has 2A,3B and 4P.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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by anuprajan5 » Thu Nov 01, 2012 12:46 am
Probability that the florist does not have to change the bouquet = 1 - (probability that the two flowers A's + probability that the two flowers are B's + probability that the two flowers are P's)
= 1 - [(2/9 * 1/8) + (3/9 * 2/8) + (4/9 * 3/8)]
= 1 - [20/72]
= 1 - 5/18
= [spoiler]13/18[/spoiler]
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by theachiever » Thu Nov 01, 2012 1:12 am
Hello Anup Thanks I have got the basic idea.2/9 is the ways of selecting the flowers from the given set of flowers which is clear.

Are the values 1/8,2/8,3/8 based on without replacement concepts?
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by anuprajan5 » Thu Nov 01, 2012 4:42 am
1/8 is the method of selecting 1 A from 8 A's.

Once you select one A from 9 flowers (2/9), then the total number of flowers is only 8. And then you select the last A from the remaining 8 ie; 1/8

This is basically probabilty when you have no replacement, as you are not replacing the 1st A back into the mix.

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by Brent@GMATPrepNow » Thu Nov 01, 2012 6:58 am
theachiever wrote:A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts 2 flowers together at random in a bouquet. However, the customer calls and says that she does not want 2 of the same flower. What is the probability that the florist does not have to change the bouquet?
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

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(diff colors) = 1 - P(same color)
= 1 - 5/18
= [spoiler]13/18[/spoiler]

Cheers,
Brent
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by YTarhouni » Sun Sep 03, 2017 1:39 pm
Alternative:
P( A&B or B&A)=2*2/9*3/8=12/72
P(A&P or P&A)=2*2/9*4/8=16/72
P(B&P or P&B)=2*3/9*4/8=24/72
Total probability=12/72+16/72+24/72=52/72=13/18
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