A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts

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BTGModeratorVI: Legendary Member; Posts: 1223; Joined: Sat Feb 15, 2020 2:23 pm; Followed by:1 members

A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts

by BTGModeratorVI » Fri Aug 14, 2020 1:15 pm

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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?

A. 5/18
B. 13/18
C. 1/9
D. 1/6
E. 2/9

Answer: B
Source: Manhattan prep

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Source: Beat The GMAT — Problem Solving | Fri Aug 14, 2020 1:15 pm

sjain296: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Fri Jun 28, 2019 9:57 pm

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts

by sjain296 » Sat Aug 15, 2020 6:51 am

Prob that florist doesn't have to change the bouquet (X) = 1 - florist has to change the bouquet (Y)

Probability of getting 2 same flowers (Y) = (2C2 + 3C2 + 4C2) / 9C2 = (1+ 3 + 6) / 36 = 10/36

Req Probability (X) = 1 - (10/36) = 26/36 = 13/18

Option B should be the answer

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts

by Brent@GMATPrepNow » Sat Aug 15, 2020 8:03 am

BTGModeratorVI wrote: ↑
Fri Aug 14, 2020 1:15 pm
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?

A. 5/18
B. 13/18
C. 1/9
D. 1/6
E. 2/9

Answer: B
Source: Manhattan prep

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

Well, 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(diff colors) = 1 - P(same color)
= 1 - 5/18
= 13/18

Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com