# The probabilitiy of selecting both candies of a different color..

#### fjvincen

##### New member
A jar has 17 red, 25 orange, and 37 blue candies. Two candies are drawn without replacement.

Find the probability that both candies will be different colors.

I've already found the probability that it will be, red then blue/blue then orange/ or both blue. So I know how to do these type of problems, but this one is giving me an issue because I'm not sure what formula to use. There are 79 total, but the number of candies aren't equal so there isn't a .33 chance of choosing a different candy each time, because as I said the probability isn't equal to that.

Any ideas?

#### ksdhart

##### New member
Well, you're very close to the right method. You're correct that the chance of drawing any color is not 33% (1/3), but you should know what it is. Let's begin by looking at the scenario where you draw a red marble, then an orange marble. Start by determining the chance of your first pick being a red marble. As a hint, if you had 5 candies in a bag, 2 of which were chocolate, and you drew one at random, what's the chance you'd get a chocolate? Then, noting that, because the picks are done without replacement, you have one less marble in the jar, what's the chance of your second pick being an orange marble? Then repeat the logic for all the other scenarios where you can have two different colors.

#### pka

##### Active member
A jar has 17 red, 25 orange, and 37 blue candies. Two candies are drawn without replacement.
Find the probability that both candies will be different colors.
There is a much easier way of thinking about this.
Some wag has named it the backdoor approach. Calculate what you do not want.

What is the number $$\displaystyle \dfrac{17\cdot 16+25\cdot 24+37\cdot 36}{79\cdot 78}~?$$ What if we subtract that from one?