probability model

Question

a bin contains 4 apples and 3 oranges. 5 pieces of fruit are selected at random Without replacement

A.) construct a probabilistic model for this experiment using combination

b.) What is the probability that the sample contains more apples than oranges?

C.) What is the probability that it contains atleast one orange?

d.) What is the expected number of oranges?

Answer 1

There are ₇C₅ = 7! / (5!(7-5)!) = 7*6/2 = 21 possible ways to choose 5 pieces of fruit from 7 pieces of fruit.  If the 5 pieces are selected at random then each of these 21 outcomes is equally likely.  

 

To get more apples than oranges you must select either 4 apples and 1 orange, or 3 apples and 2 oranges.  There are 3 ways to select 4 apples and 1 orange (you must select all 4 apples and you can select any 1 of the 3 oranges), and there are 4*3 = 12 ways to select 3 apples and 2 oranges (you can not select any 1 of the 4 apples in 4 ways, and you can not select any 1 of the 3 oranges in 3 ways).  So there are 3 + 12 = 15 total ways to get more apples than oranges, so the probability that this occurs is 15/21 = 5/7 ~ 0.714 

 

There is no way to select 5 pieces of fruit without selecting at least one orange, so the probability that the selection contains at least one orange is 1 

 

To find the expected number of oranges you must find the probability of getting each number of oranges.  We already know 

 

P(0 oranges) = 0

P(1 orange) = 3/21

P(2 oranges) = 12/21

 

Since the probabilities must add up to 1 we must have P(3 oranges) = 6/21.  You could also figure this out by noting that in order to get 3 oranges you must select all 3 oranges (in 1 way) and any 2 of the 4 apples (in ₄C₂ = 6 ways).  

 

Then E[# oranges] = sum from 0 to 3 of P(#oranges)*#oranges 

 

= 0*0 + (3/21)*1 + (12/21)*2 + (6/21)*3

= 0 + 3/21 + 24/21 + 18/21

= 45/21 = 15/7 ~ 2.143