Probability

jenlee wrote: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?

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

vipulgoyal wrote:experts please suggest whats wrong with this approach

2/9 * 3/8 + 2/9 * 4/8 + 3/9 * 4/8 = 13/36

It would help if you explained what each value stands for.

Having said that, it looks like you're trying to consider all possible ways in which we can have different flowers. For example, I believe 2/9 * 3/8 stands for the possibility that we select an azalea and then a buttercup. This part is correct P(select azalea then buttercup) = 2/9 * 3/8 = 6/72.

However, your solution also needs to include P(select buttercup then azalea) = 3/9 * 2/8 = 6/72.

In fact, all three parts of your solution are missing the second option.

So, if we double your answer, we get the correct response.

Cheers,
Brent