Probability Problem Sets
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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?
- anuprajan5
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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]
= 1 - [(2/9 * 1/8) + (3/9 * 2/8) + (4/9 * 3/8)]
= 1 - [20/72]
= 1 - 5/18
= [spoiler]13/18[/spoiler]
Regards
Anup
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Anup
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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?
Are the values 1/8,2/8,3/8 based on without replacement concepts?
- anuprajan5
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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.
Regards
Anup
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.
Regards
Anup
Regards
Anup
The only lines that matter - are the ones you make!
https://www.youtube.com/watch?v=kk4sZcG ... ata_player
Anup
The only lines that matter - are the ones you make!
https://www.youtube.com/watch?v=kk4sZcG ... ata_player
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First, we can rewrite the question as "What is the probability that the two flowers are different colors?"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?
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