Question

I DO NOT JUST WANT THE ANSWER
i want to learn

can anyone give me a formula for this question and a brief steps. HELP PLEASE!

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. The probability that you have 1 cherry candy and 2 lemon candies

Answer 1

Well, that's good. You don't have to put it in all-caps, though.

So, here:

The probability of an event \(E\) is:
\[
P(E)=\frac{\text{desired outcomes}}{\text{possible outcomes}}
\]The probability of multiple, independent events \(A,B,...\) all happening (that is, the outcome of one event does not change the probability of the other) is:
\[
P(A\cap B\cap\cdots)=P(A)\cdot P(B)\cdots
\]So, we are going to have that (1) the probability of picking a cherry for your first time is:
\[
P(A)=\frac{6}{6+3+2}=\frac{6}{11}
\]The next time you pick, you now are missing one candy from the bag, so we have, for one lemon:
\[
P(B)=\frac{2}{5+3+2}=\frac{1}{5}
\]Now, we're missing one lemon and a cherry, so our final pick will have a probability of:
\[
P(C)=\frac{1}{5+3+1}=\frac{1}{9}
\]Now, we wish to find the probability of ALL of these events, that is:
\[
P(A\cap B\cap C)=P(A)\cdot P(B)\cdot P(C)
\]What would this be?

Answer 2

6/11*2/10*1/9=12/990=0.01212=1.2%?

Answer 3

That is correct.

Answer 4

thank you!