>>Dr. Holmes: The head coach of the football team believes that Jones should be the starting
quarterback for the game this week. Eight of the other coaches agree with the head coach,
but the remaining seven prefer other choices. A committee of five of the fifteen coaches
is formed to make a recommendation. What is the probability that of five of the fifteen
coaches selected at random for the committee to make the decision, at least four members
of the committee will agree with the head coach?
We approach this problem first by defining the random variable. What are we interested
in? Answer: number of coaches that agree with the head coach.
We will call this random variable X. Number of coaches that agree with the head coach.
What type of random variable is it: only two choices here, continuous and discrete. Answer:
discrete because the random variable is measured in whole numbers.
Next, what values can the random variable take on, meaning what are the possible values
of x. Answer: zero, one, two, three, or four or
five, because there's only five members on the committee.
Finally: what values of the random variable are we interested in finding a probability for?
The answer here is actually two values, the probability that X is equal to four or
X is equal to five. And this is because we asked the question “at least four” which of
course includes five.
Now we have to count: first, how many unique ways can we select four members from the group
of coaches who agree with the head coach?
Here the combinatorial formula provides us with the answer:
eight combinatorial four is a total number of unique ways that we canselect four from the people who agree.
Now, we have one more slot to fill because we’re gonna have five in our committee and
those have to come from the seven who disagree with the coach. And that’s gonna be combinatorial,
seven combinatorial one. We only have one more step to go and that is, how many total
possible five member committees can you create from fifteen unique elements. And the answer
there is fifteen combinatorial five, okay? This is the total number of unique ways to
create a committee of size five from fifteen possible members.
By just using counting formulas we have now created something that is called the Probability
Density Function. This one is called the Hypergeometric Probability Density Function.
Probability density functions are mathematical formulas that describe exactly the probability
of certain experiments. We can use these formulas to calculate probabilities if the experiment
follows the mathematical requirements of the Probability Density Function. These things
are truly amazing as we discussed earlier..
Probability density functions, however, are only useful for specific experiments. In this
case of selecting a committee of coaches the experiment tells us that the Hypergeometric
Probability Density Function is appropriate. The rules are: the population must be divisible
into two and only two independent subsets (Coaches who agree and those who do not, in
this example); the experiment must have changing probabilities of success with each trial (the
fact that the names selected to be on the committee cannot be selected twice
in this example makes this true here) this is basically saying we draw without replacement;
the last the random variable must be discrete, rather than continuous.
Now to the solution:
We have one more step, the question is what is the probability that the committee has
four OR MORE coaches who agree. Now we are interested in not only the probability of
having exactly four people, which we have already here, but also the probability of
having five on the committee who agree. We will have to solve the problem twice - one
for each number and then add the two probabilities together. We know this because the addition
rule reminds us that the word OR translates to addition in terms of the addition rule.
The general formula for the hypergeometric distribution is as follows.
The probability of X is equal to A combinatorial X, N minus A, combinatorial to n minus x,
divided by capitol N combinatorial small n. where capitol N is the total number of items we're
drawing from, in this case, the coaches. A is the number of those that we’re interested in.
These are coaches who agree. And x is the number that we’re interested in for
finding the probability. To finally solve the problem, we simply add on… this is going to be
eight coaches who agree, but this time we want five, combinatorial five. There are
seven who disagree and we want none, because all five of them are agreeing and we divide
that by fifteen combinatorial five, okay? When we want to solve that we can work it
out and the answer’s gonna be point eighteen eighteen. Tadaa.