The question is from a standardized test. It is based around a type of probability. I have not taken a class that involves probability, as it is not taught in our standard curriculum. I am not positive on if my partial efforts are correct, but I and many others I know would appreciate some clarification. The problem goes (keep in mind this is from memory, I do not have answer choices memorized. So if you are unsure I cannot provide any choices to go from.)
"You have a jar of 10 cookies. In this jar are 5 chocolate chip cookies, 3 oatmeal cookies, and 2 raisin cookies. You reach in at random and take out a cookie. Without replacing the first cookie you pulled, you pull out a second. What is the probability that the first and second cookies will be the same type?"
As stated, this is from memory. I have spoken with friends to confirm and all agree this is the exact problem. We may be wrong, so if this is mathematically impossible to solve, be sure to let me (us) know.
From my very limited knowledge of probability, I cannot see a way to form one probability from three. You would take the probability of the first pick (for each type of cookie), 5/10, 3/10, and 2/10, and multiply them respectively by 4/9, 2/9, and 1/9. That would give you the probability of picking the same for each type.
How would you form one single probability from those three (if possible)?
As I said. I could be wrong, as if this is just not possible, then I obviously am not recalling it clearly enough. Thanks in advance.
The answer is to sum them. The probability for selecting 2 cookies of the same type from a choice of the three types when selecting any 2 from all 10 cookies, is probability for a union of three disjoint events: each choice of type. So the probability is an addition of their individual probabilities.
That is: $\dfrac 5{10}\cdotp\dfrac 49+\dfrac 3{10}\cdotp\dfrac 29+\dfrac 2{10}\cdotp\dfrac{1}{9}$ also written as: $\Big({\dbinom 52+\dbinom 32+\dbinom 22}\Big)\div{\dbinom {10}2}$
Which evaluates to $\dfrac{14}{45}$
That is all.