5. A bin contains 5 apples and 7 oranges. 6 pieces of fruit are selected at random.
This problem could be done using a tree diagram, but if there are 6 drawings, and if on each drawing there are two choices for the type of fruit, then there would be 2^{6} = 64 branches at the end of the tree diagram, which more bit ungainly than is preferable for instructional purposes. So we will use combinations instead.
a) If the pieces of fruit are selected at random, then any sample is as likely as any other sample, so we use an equally likely probabilistic model. In an equally likely probabilistic model, if E is an event and S is the sample space then the probability of the event
Since the experiment is choosing 6 pieces of fruit out of a basket of 12 pieces without replacement, the number will be given be either permutations of combinations.In this case we do not care about the order in which the pieces are selected, only in the number of apples and oranges, so we can use combinations.
In this problem we will need to compute the probabilities of all possible numbers of apples from 0 to 5 or 6. To get the number of each type of sample we use a process similar to the computations of the lottery hands.
Number of apples 


Number of Samples 
0 

1 x 7 = 
7 
1 

5 x 21 = 
105 
2 

10 x 35 = 
350 
3 

10 x 35 = 
350 
4 

5 x 21 = 
105 
5 

1 x 7 = 
7 
6 

0 x 1 = 
0 
Totals 


924 
Notice that it is impossible to get a sample which contains 6 apples if there are only 5 apples in the bin. That is reflected in the fact that if you compute a combination number using the formula with r consecutive factors on top and bottom starting with an n on the top and an r on the bottom, if r is bigger than n then there will be a factor of 0 on the top of the fraction, and the answer will be 0. We convert these numbers to probabilities.
Number of apples 
probabilities 

0 
7/924 = 
1/132 
1 
105/924 = 
15/132 
2 
350/924 = 
50/132 
3 
350/924 = 
50/132 
4 
105/132 = 
15/132 
5 
7/924 = 
1/132 
6 
0/924 = 
0 
Totals 

1 
b) With this probabilistic model, we can answer any of these questions. To get the probability that there are more apples than oranges, we find the outcomes in which this event happens, and add up the probabilities of all the outcomes.
Number of apples 
Probabilities 
4 
15/132 
5 
1/132 
6 
0 
Totals 
16/132 = 4/33 
Since the bin has more oranges than apples, as we would expect, the probability of getting more apples than oranges is less than 1/2.
c) To get the probability of at least one orange, add up the probabilities of all the outcomes in the event
Number of apples 
Probabilities 
0 
1/132 
1 
15/132 
2 
50/132 
3 
50/132 
4 
15/132 
5 
1/132 
Totals 
1 
This is a trick question. If there are only 5 apples in the bin, then if you have 6 pieces of fruit, at least one of them has to be an orange. Since it is certain that there will be at least one orange, the probability of this event it 1. That is the same result we get from using the probabilistic model.
Often there will be a quick easy way to compute probabilities, and the question comes up, "When will the quick and easy way work?" The answer to that question is that if the quick, easy way gives you the same answer as using the probabilistic model, then the quick, easy way will work. If the quick, easy way gives you a different answer then you would get using the probabilistic model, then the quick, easy way will not work.
d) To compute the expected number of apples, we multiply the number of apples in each outcome times the probability of the outcome and add up the products
Number of apples 
Probabilities 
Number x Probability 
0 
1/132 
0 
1 
15/132 
15/132 
2 
50/132 
100/132 
3 
50/132 
150/132 
4 
15/132 
60/132 
5 
1/132 
5/132 
6 
0 
0 
Totals 

330/132 = 2.5 
We would expect to get two and a half apples on the average.
In this case there is a quick and easy way to get the answer. If 5/12 of the pieces of fruit in the bin are apples, then 5/12 of the sample should be apples. 5/12 of 6 is 30/12 = 5/2 = 2.5. Here the quick, easy way gives us the same answer that using the probabilistic model does. While it is not really apparent that this method will always work, it is possible to prove that it will. The proof, however, is not quick and easy.