SOLUTION: Suppose that a committee of two boys is being chosen at random from the five boys Al, Bill, Carl, Dan, and Eli. Find the probability that Dan and Eli are both on the committee

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Question 1157530: Suppose that a committee of two boys is being chosen at random from the five boys Al, Bill, Carl, Dan, and Eli. Find the probability that Dan and Eli are both on the committee
Found 3 solutions by KMST, ikleyn, n2:
Answer by KMST(5403) About Me  (Show Source):
You can put this solution on YOUR website!
You want to find all the combinations of 2 names taken from the 5 different names given.
There are names and a formula related to that, and in order to communicate with the teacher and others,
it would be convenient to at least know the names.
You may use logical thinking to do the calculations and not need the formula.
If you have 5 names in a hat, there are 5 different possibilities when picking one.
When picking a second name, there will be 4 names in the hat, so for each of the 5 possible first picks, there are 4 possibilities for the second pick.
So, here are 5%2A4=20 possible ordered sets of 2 names taken from a set of 5 such items.
There are 20 ordered pairs of names possible, so we say that there are 20
permutations of 2 item taken out of a set of 5 items.
In this case they did not ask for a two people and their particular roles in the committee, so you are looking for a number of sets of two people and the order you list them does not matter.
Any ordered set of 2 names can be chosen in {{[2}}} different orders.
For example, you can get (Al, Bill) if Al is the first name picked, but you can also pick Bill first and get the ordered set (Bill, Al).
In this case, you do not care about the order the names are drawn, and know that each set is found twice, listed in different order, in the 20 possible results of drawing one number after the other.
That means that in those 20 results there are just 20%2F2=highlight%2810%29 sets of 2 names.
That is the number of equally probable possible different sets or combinations of 2 names you can get out of the set of 5 available different names.
That is 10 possible different committees, all equally probable when choosing at random.
The set Dan%2C+Eli is 1 of the 10 possible results when picking names at random. The probability of getting a committee of 2 made by Dan and Eli selecting at random is highlight%281%2F8%29

THE FORMULAS AND MORE NAMES:
If order matters, you are looking for the number of highlight%28permutations%29 you can make.
If order does not matter, you are looking for highlight%28combinations%29 .
The number of permutations of n objects made from n different items is n%2A%28n-1%29%2A%28n-2%29%2A%22...%22%2A3%2A2%2A1 , and.
That product is usually written including that last factor, even though multiplying times 1 does not change the result.
The product of the first n counting numbers is called highlight%28factorial%29 of n , and its symbol is highlight%28%22n%21%22%29 .
The number for permutations from a set of n different items, taking a smaller number r%3Cn of them is represented by different symbols and is the product of only r factors.
It symbols are P%28n%2Cr%29 , %22+%22%5EnP%5Br%5D , or nPr and its formula is usually given using factorials, as

If order of the r items does not matter, the number of sets of r items that can be made from n different items is combinations from n items taken r at a time, symbolized as
C%28n%2Cr%29 , %22+%22%5EnC%5Br%5D , nCr , or %28matrix%282%2C1%2Cn%2Cr%29%29
Knowing that each group of r items appear r%21} times in different orders in P%28n%2Cr%29 , we get the formula for C%28n%2Cr%29 from P%28n%2Cr%29%2Fr%21 , as
highlight%28C%28n%2Cr%29=%28matrix%282%2C1%2Cn%2Cr%29%29=n%21%2F%28%28n-r%29%21r%21%29%29

Answer by ikleyn(53955) About Me  (Show Source):
You can put this solution on YOUR website!
.
Suppose that a committee of two boys is being chosen at random from the five boys
Al, Bill, Carl, Dan, and Eli. Find the probability that Dan and Eli are both on the committee
~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution by @KMST is very wordy and her final answer   ( 1%2F8  for the probability )   is incorrect.
        I came to provide a correct short solution.


In this problem, each committee is (unordered) set of two boys, randomly selected from 5 boys.


So, in this problem, we work with COMBINATIONS.


For 5 boys, there are  C(5,2) = %285%2A4%29%2F2 = 10 different combinations / (ten possible different committees).


Of them, only one (unordered) pair (Dan, Eli) is favorable.


Therefore, the desired probability is  P = favorable%2Ftotal = 1%2F10 = 0.1 = 10%.


ANSWER.  The desired probability is  P = 1%2F10 = 0.1 = 10%.

Solved correctly.



Answer by n2(93) About Me  (Show Source):
You can put this solution on YOUR website!
.
Suppose that a committee of two boys is being chosen at random from the five boys
Al, Bill, Carl, Dan, and Eli. Find the probability that Dan and Eli are both on the committee
~~~~~~~~~~~~~~~~~~~~~~~~~~~


In this problem, each committee is (unordered) set of two boys, randomly selected from 5 boys.


So, in this problem, we work with COMBINATIONS.


For 5 boys, there are  C(5,2) = %285%2A4%29%2F2 = 10 different combinations / (ten possible different committees).


Of them, only one (unordered) pair (Dan, Eli) is favorable.


Therefore, the desired probability is  P = favorable%2Ftotal = 1%2F10 = 0.1 = 10%.


ANSWER.  The desired probability is  P = 1%2F10 = 0.1 = 10%.

Solved.