A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts two flowers together at random in a bouquet. However,

VJesus12 wrote: ↑

Thu Sep 03, 2020 6:29 am

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

Solution:

Calculating the probability that the florist does not have to change the bouquet is the same as calculating the probability that the bouquet has two different flowers. We can use the following formula:

1 = P(getting two of the same flower) + P(not getting two of the same flower)

Let’s determine the probability of getting two of the same flower.

P(2 azaleas) = 2/9 x 1/8 = 2/72

P(2 buttercups) = 3/9 x 2/8 = 6/72

P(2 petunias) = 4/9 x 3/8 = 12/72

The probability of getting two of the same flower is, therefore, 2/72 + 6/72 + 12/72 = 20/72 = 5/18.

Thus, the probability of not getting two of the same flower is 1 - 5/18 = 13/18.

Alternate Solution:

Without any restrictions, there are 9C2 = 9!/(2!*7!) = (9*8)/2 = 36 ways to pick two flowers from a total of 9 flowers.

The two azaleas can be chosen in 2C2 = 1 way, two buttercups can be chosen in 3C2 = 3 ways and two petunias can be chosen in 4C2 = 4!/(2!*2!) = (4*3)/2 = 6 ways. In total, there are 1 + 3 + 6 = 10 ways to pick two flowers of the same kind and thus, 36 - 10 = 26 ways to pick two bouquets that contain two different kinds of flowers. We see that the probability that the selection of two flowers does not contain the same kind of flowers is 26/36 = 13/18.

Answer: B