We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
221
3
avatar+85 

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

 Feb 25, 2019
 #1
avatar+85 
0

the answer is not 1/5 btw

 Feb 26, 2019
 #2
avatar+5226 
0

\(\text{3 oranges or 6 apples are mutually exclusive events so}\\ P[\text{3 oranges} \cup \text{6 apples}] = P[\text{3 oranges}]+P[\text{6 apples}]\)

 

\(\text{There are a total of }n=\dbinom{8+3-1}{3-1}=45 \text{ ways of choosing the 8 fruits}\\ \text{Select 3 oranges. We now have 5 pieces of fruit to choose from either banana or apple}\\ \text{there are }\dbinom{5+2-1}{2-1}=6 \text{ ways this can be done, with a probability of }\dfrac{6}{45}=\dfrac{2}{15} \text{Select 5 apples. We now have 3 pieces of fruit to choose from either orange or banana}\\ \text{There are }\dbinom{3+2-1}{2-1}=4\text{ ways of doing this with a probability of }\dfrac{4}{45}\\ P[\text{3 oranges}\cup \text{6 apples}] = \dfrac{2}{15}+\dfrac{4}{45} = \dfrac{2}{9}\)

.
 Feb 26, 2019
 #3
avatar+85 
+3

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. 

 Feb 28, 2019
edited by alskdj  Mar 1, 2019

7 Online Users

avatar