# Need guidance on solution - Probability/Counting

• Oct 9th 2012, 03:16 PM
yvonnehr
Need guidance on solution - Probability/Counting
(Itwasntme)Among recent graduates of mathematics departments, half intend to teach high school. A random sample of size 2 is to be selected from the population of recent graduates.
(a) If mathematics departments had only four recent graduates total, what is the chance that the sample will consist of two graduates who intend to teach high school?

What solution method does this call for?
• Oct 9th 2012, 05:46 PM
johnsomeone
Re: Need guidance on solution - Probability/Counting
The numbers are small enough that, although it isn't required, you can directly enumerate the possibilities. This can prove enlightening.

The general idea is:
1. All possible ways to choose 2 grads out of those 4 grads is the sample space.
2. Find out how many elements (pairs of students) in that sample space correspond to having choosen 2 grads who intend to teach high school.

Though these aren't the numbers for this problem, but suppose the sample space had 500 pairs of students. And suppose what you were seeking corresponded to 100 of those possible pairs. Then the probability that you found what you were seeking would be 100/500 = 20%.
• Oct 9th 2012, 09:18 PM
yvonnehr
Re: Need guidance on solution - Probability/Counting
Based on what you said I used the combination formula. But I get 6, not 1/6. I'm sorry, but I don't understand how to apply your example to this problem. Can you explain further?
• Oct 10th 2012, 06:32 AM
johnsomeone
Re: Need guidance on solution - Probability/Counting
If you know where the 6 came from, you've done the harder half of the problem - you're almost there. I'll work it out in concrete detail so you can "see" what's going on.

Suppose the population of graduates is {Mary, Sue, Bob, Pat}. Suppose Mary and Bob are the two that intend to teach high school.
When you randomly pick two of those four (a random sample of 2), you'll get, with equal probability, one of these six outcomes:

Pair #1 = {Mary, Sue}
Pair #2 = {Mary, Bob}
Pair #3 = {Mary, Pat}
Pair #4 = {Sue, Bob}
Pair #5 = {Sue, Pat}
Pair #6 = {Bob, Pat}

In other words, by picking a random sample of two of them, you've randomly selected one of those six pairs. Since Mary and Bob are the future teachers, what's the probability that you picked the pair that intend to teach high school? It's the probability that you picked Pair #2 = {Mary, Bob} out of those six possibilities. That's 1 chance out of 6 possibilities, so the probability of it happening is 1/6.

It's *exactly* analogous to this: What's the probability that, when you roll a ordinary six-sided die, that your roll is a 2? It's 1/6. Why? Because there are 6 equally likely outcomes, and rolling a 2 happens in only one of them.

Here are 8 more problems (I'll do the first two. You need only understand probability and look at that list to get the answer - no work is required.):

1) Assuming Pat is a man, what's the probability that you picked two women?
(On the list of outcomes, two women occurs only for Pair #1 = {Mary, Sue}. The probability of picking that one out of those 6 is, again, 1/6.)

2) Assuming Pat is a woman, what's the probability that you picked two women?
(On the list of outcomes, two women occurs only for Pair #1 = {Mary, Sue}, Pair #3 = {Mary, Pat}, and Pair #5 = {Sue, Pat}. We'll have picked two women if we've selected any of those 3 pairs out of the 6 possible pairs. Thus the probability is 3/6 = 1/2.)

3) Assuming Pat is a man, what's the probability that you picked at least one woman?

4) Assuming Pat is a woman, what's the probability that you picked at least one woman?

5) What's the probability that Bob is a member of the pair you picked?

6) What is the probability that either Bob or Sue, or both, is a member of the pair you picked?

7) What is the probability that either Bob or Sue, but not both, is a member of the pair you picked?

8) What is the probability that Mary isn't a member of the pair you picked?
• Oct 10th 2012, 06:42 AM
yvonnehr
Re: Need guidance on solution - Probability/Counting
I get it!!! Thank you!!! :-D
• Oct 10th 2012, 09:21 PM
yvonnehr
Re: Need guidance on solution - Probability/Counting
Hey John,
So I really like the explanation and additional exercises you gave me. They were extremely helpful. The next part of the question, however, uses a much larger number with which I can't write out all the possible combinations. See below. Can you help me with this?

***If mathematics departments had 10 million recent graduates, what is the chance that the sample will consist of two graduates who intend to teach high school?

I don't even want to tell you what I tried, because the answer I got was ridiculous. :-/
• Oct 10th 2012, 09:31 PM
yvonnehr
Re: Need guidance on solution - Probability/Counting
Wake up, John. It's not midnight yet.
• Oct 11th 2012, 08:43 AM
johnsomeone
Re: Need guidance on solution - Probability/Counting
OK - you'll obviously need a different approach if there are a million graduates, of which half intend to teach high school. The problem you did with 4 graduates is still the thing to always keep in mind, because although you can't *numerically* solve problem with that approach, that's still *conceptually* how you need to think about every problem. It's still exactly "what's going on" - it's just that now the size of that list is too huge to be directly useful.

So here's the different approach. I'll do it with the original problem, and then let you consider your 1,000,000 graduates problem:
In the original problem we wanted to find the probability of randomly picking the two future high school teachers (Mary and Bob) from the pool of four graduates (Mary, Sue, Bob, Pat). We solved it by looking at the number of outcomes that satisfied our requirements, versus all possible outcomes. In particular, the act of "picking two at random" was treated as being done "at once". However, the act of "picking two at random" can also be treated as "picking one at random, and then, from those that remain, picking another one at random".

The distinction between "picking two at random at once" versus "picking one at random, and then, from those that remain, picking another one at random" goes to a very common distinction in probability about whether we care about the sequencing and order of our results or not. In the way we originally did the problem, there was no such attention given to sequencing and order - and that was fine, and we did it correctly that way. And conceptually, there is no sequencing and order to that problem. However, we could've treated as though it had sequencing and order. We could've treated it as "picking one at random, and then, from those that remain, picking another one at random".

&&&&&&&&&&&&&&&&&&&&&&&&&

Before doing that, you'll need to understand about the phrase "AND THEN" in probability. When you do one thing AND THEN do another, the probability that both happened is the product of the probabilities that each happened (there's a little proviso here that I'll ignore for now). Consider the probability of flipping 3 heads in a row with a fair coin. Three flips of a fair coin has the following possible outcomes, all equally likely. (Note that the order of the flips is included.)
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (8 possible outcomes)
Thus the probability of flipping 3 heads is one (the "HHH") out of 8 = 1/8.
(Likewise, the probability of flipping exactly 2 heads (the "HHT", "HTH", and "THH") is 3 out of the 8 possible outcomes, so Prob = 3/8. Etc.)
However, "3 heads" is the same as "Heads AND THEN heads AND THEN heads", so we multiply the probability of each of those individual flips being heads together:
Probability = (prob that 1st flip is heads) $\times$ (prob that 2nd flip is heads) $\times$ (prob that 3rd flip is heads).

So Probability of 3 heads $= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

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Back to the original graduates problem:
Using "picking one at random, and then, from those that remain, picking another one at random", we want to find the probability that we end up picking the future teachers (Mary and Bob) out of all the graduates (Mary, Sue, Bob, Pat).
That means 1st pick Mary or Bob, AND THEN pick the other one out of the remain pool.
At the start, for our 1st pick, we have 4 possible outcomes (Mary, Sue, Bob, Pat) of which 2 satisfy our requirement that we picked Mary or Bob.
Thus the probability of picking Mary or Bob with the first pick is 2/4 = 1/2.
So now imagine we've choosen Mary or Bob with the 1st pick. Since we don't know which of those two it was, I'll call the other one TheOtherOne. Thus if we chose Mary, TheOtherOne = Bob, and if we chose Bob, TheOtherOne = Mary.
For the second pick, we want to pick TheOtherOne out of those that remain: (Sue, Pat, TheOtherOne). That's 1 possible outcome out of 3 possibilities, so the probabiltiy of picking TheOtherOne out of those that remain after the 1st pick is 1/3.
Thus the probability of "picking Mary or Bob, AND THEN, from those that remain, picking the other one", is

Prob(1st = Mary or Bob, AND THEN 2nd = the other one)

= Prob(1st = Mary or Bob) TIMES Prob(2nd = the other one)

$= \frac{2}{4} \times \frac{1}{3} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.

And we know that's the right answer.

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Another example:
Question:
You have 60 jelly beans in jar: 40 blue and 20 red. You reach in and eat one, then reach in and eat another. What's the probability that you ate two blue jelly beans?
Prob(1st pick is blue AND THEN 2nd pick is blue)

= Prob(1st pick is blue) TIMES Prob(2nd pick is blue, given that you already ate the first blue pick)

$= \frac{40}{60} \times \frac{39}{59} = \frac{2}{3} \times \frac{(3)(13)}{59} = \frac{(2)(13)}{59} = \frac{26}{59} = .441 = 44.1\%$.

Do you see why Prob(2nd pick is blue, given that you already ate the first blue pick) = 39/59? After removing 1 blue jelly bean, there are 59 jelly beans left, of which 39 are blue, and 20 are red. So picking a blue one then has 39 winning outcomes out of 59 equally likely outcomes.

&&&&&&&&&&&&&&&&&&&&&&&&&

If you understood that jelly bean example, then you should be ready to solve your problem about 1,000,000 graduates, half of which intend to be teachers.
• Oct 11th 2012, 06:08 PM
yvonnehr
Re: Need guidance on solution - Probability/Counting
Got it! Here's how the problem is set in my mind after reading your lesson...

I have 10 million graduates, 5 million of them that intend to teach. I pick a sample of 2 from those 10 million graduates. What are the odds that those 2 intend to teach?

The probability of the first one picked intending to teach is 5 million/10 million = 1/2. The probability of the next one being picked intending to teach is 4,999,999/9,999,999 = .499.... So the odds of my sample of 2 intending to teach is 1/2 X .499... = .2499...

Hooray!!! Thank you! :-D I'm a happy camper.
• Oct 12th 2012, 09:19 AM
johnsomeone
Re: Need guidance on solution - Probability/Counting