+88 Pat wants to select 8 pieces of fruit to bring in the car for the people he's driving to Montana with. He randomly chooses each piece of fruit to be an orange, an apple, or a banana. What is the probability that either exactly 3 of the pieces of fruit are oranges or exactly 6 of the pieces of fruit are apples?
THANKS FOR ALL HELP
+6251 \(\text{3 oranges or 6 apples are mutually exclusive events so}\\ P[\text{3 oranges} \cup \text{6 apples}] = P[\text{3 oranges}]+P[\text{6 apples}]\)
\(\text{There are a total of }n=\dbinom{8+3-1}{3-1}=45 \text{ ways of choosing the 8 fruits}\\ \text{Select 3 oranges. We now have 5 pieces of fruit to choose from either banana or apple}\\ \text{there are }\dbinom{5+2-1}{2-1}=6 \text{ ways this can be done, with a probability of }\dfrac{6}{45}=\dfrac{2}{15} \text{Select 5 apples. We now have 3 pieces of fruit to choose from either orange or banana}\\ \text{There are }\dbinom{3+2-1}{2-1}=4\text{ ways of doing this with a probability of }\dfrac{4}{45}\\ P[\text{3 oranges}\cup \text{6 apples}] = \dfrac{2}{15}+\dfrac{4}{45} = \dfrac{2}{9}\)
.
+88 Nevermind guys. That's wrong too.
It is impossible for Pat to select both 3 oranges and 6 apples, so we can calculate the probabilities of these mutually exclusive cases separately and then add to get our final answer. The probability that 3 particular pieces of fruit will be oranges and the rest will not be is given by (1/3)^3(2/3)^5=32/6561, and there are C(8,3)=56 ways of selecting three pieces of fruit to be the oranges so the probability that 3 will be oranges is 56*(32/6561)=1792/6561. Similarly, the probability that 6 particular pieces of fruit will be apples and the other two won't be is given by (1/3)^6*(2/3)^2=4/6561 and there are C(8,6)=28 ways of selecting which ones will be the apples, so multiplying again gives us a probability of 28*(4/6561)=112/6561. Adding those two probabilities give us our final answer: 1792/6561+112/6561=1904/6561.
