Word Translations Probability Chapter5

[msriniva]

Pg 101 Q#10 in the problem set. The answer stated in the guide is 13/18.

The answer according to my calculation is 7/9.
A florist has 2A, 3B and 4P's. Se puts 2 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?

My solution
P(A) 2 times is 2/10*1/9
p(B) 2 times is 3/10*2/9
p(P) 2 times 4/10*3/9

hence, P(A) or P(B) or P(C) 2 times is 1/45+1/15+2/15.
Therefore P(not 2 flowers) 1-(2/9)=7/9

Please let me know if I am missing anything here.

Thanks
Mani

[suhas_rg]

PROB(2 flowes selected not same) = 1 - PROB(2 flowes selected are same)
X' = 1 - X
X' = 1 - (2C2 + 3C2 + 4C2)/9C2
X' = 1 - (2!/2!0! + 3!/2!1! + 4!/2!2!)/(9!/7!2!)
X' = 1 - (1 + 3 + 6)/36
X' = 1 - 10/36 = 26/36 = 13/18

[JonathanSchneider]

To the first poster:
your approach can work, but you've made a simple mistake. You counted the total number of flowers as 10, when in fact there are only 9. If you had 9*8 as the denominator of your fractions, then everything would work out just fine.