Sunday, May 25, 2008

Types of Probability and Exclusivity

So, this is a double scribe post, which covers the lessons we did on May 22 and May 23 (Thursday and Friday respectively).

So lets get started, shall we? First off is our lesson from May 22...


So last Thursday, we talked about the two types of probability. There are dependent and independent probabilities, and thats what we'll explore here.

Slide Two:
The first thing we did was make a tree, which we've all done before. This tree displays all the possible outcomes of flipping two coins. Easy right? We know probability of each outcome is 1/4 because there are four possible outcomes, and each one is one of those four.

Slide Three:
Moving forward, we have "an entirely different question that is not at all similar to the one we
just did." Please note the quotation marks. Because, this question is very similar to the one we just did. Instead of heads and tails, there are reds and blues. But this is similar but not the same. Why? Because unlike our previous probability question, this question deals with a dependent probability, whereas our previous one dealt with an independent probability. Remember, this lesson was supposed to be about dependent and independent probabilities?

But Paul, you say, what the heck are dependent and independent probabilities anyway? Im confused!

Well, to put it simply, independent probabilities are probabilities that are affected by the steps before them. Why don't you think about it in context, like this:

If you flip a coin, and then you flip another entirely different coin, does the chance of you getting tails on the second coin change because you flipped the previous coin?

Answer? Of course not, excluding far-out possibilities such as your hand got so tired flipping the previous coin that you put less energy into flipping your second coin and gave it 0.001% better chance of landing on tails. This is an example of an independent probability, where both probabilities are, well, they're independent of each other. Makes sense, right?

However, with our second question, our probabilities are dependent. Why? Because when you choose a marble (wearing a blindfold, earmuffs, and nose plugs so you cannot possibly see, hear or smell which marble you're picking), you also
remove that marble, thus changing the probability of you getting the same colour marble when you draw a second one. This is all displayed in the slide two tree, where if you drew a red marble and removed that, your chance of drawing a red marble again the second time is 2/5, whereas if you drew a blue marble the first time, you have a 3/5 chance of getting a red marble when you draw a second time. See? They're different, which is exactly why this is a dependent variable.

And that is basically the gist of this entire lesson, so I'll just summarize the rest of the slides...

Slide Four:
Okay, something changed in this question, and I'll give you a cookie if you can spot it.

Alright, found it? Here, have a cookie. You clearly noticed that our question changed in that the marble is no longer simply discarded (or "thrown out the window" as Mr. K would say [hey, I rhymed]). Instead of removing a marble when we draw it, we just put it back. Because we put the marble back, our probability doesnt change, and it becomes independent. Now that its independent, all you'd have to do is change those R's to H's and B's to T's, and you've got your original heads and tails question! Magic right?

Which brings up another point Mr. K mentioned. A lot of the time, your question boils down to stuff much like the heads and tails question, just wrapped up in a pretty little. Remember that (say it to yourself five times, or something), and you'll find things will be a lot easier.

Slides Five & Six:
Just a rehash of what I already wrote, basically definitions of what dependent and independent variables are (in simpler, less rambling and rant-y [is that a word?] terms).

Slide Seven:
A few questions, dealing with figuring out whether or not the question is describing an independent or dependent probability. Simple, right? I'll explain why quickly.

a) Independent because the outcome of the coin toss doesn't affect the outcome of the dice roll (again, this is factoring out such insane possibilities and how your possibility might be changed if you did both at once, or something along those lines).

b) Dependent because when you draw the first card, you remove it from the deck and change the amount of cards in the deck. So when you draw your second card, your deck is one card smaller than it would be had you not removed the first card, and you can no longer draw that particular card. Thus, the first outcome affects the second outcome.

c) Independent because now your deck is static and the probabilities always remain the same (excluding time travel and such nonsense).

Slide Eight:
Slide eight takes our second slide's question one step further by adding another coin flip. Nothing terribly new here, it just makes the chances of getting a particular combination smaller. I won't bother explaining this in great detail, but I'd like to point out that the final probability of an outcome (say, HHH, or flipping three heads in a row) is equal to the product of its previous steps probabilities (in other words, 1/2 * 1/2 * 1/2 = 1/8, not a coincidence), where each coin flip had a 1/2 chance of landing Heads.

Slide Nine:
Now to follow our trend of making small changes to our previous questions, we take our old question and put a fresh splash of paint on it and volia, we get something entirely new. Fortunately for me, all I have to do is take our old solution, paste it on to our new slide and say Boy = H, Girl = T. But wait, our question says, whats the probability of mom and pop getting exactly two girls and one boy. So what do we do? We find all the answers that have two girls (T's) and one boy (H's). Since there are three possible outcomes out of 8 that have two T's and one H, we determine that the chance of getting two girls and one boy exactly is 3/8.

Slide Ten:
Ignore this, this was my poor and not at all though out attempt at solving the question, and it is completely wrong.

Slide Eleven:
Our final slide, with the correct solution, courtesy of Kristina. Another variation on our girl/boy question, except with a different amount of steps and different conditions. So for the first draw, if we draw a blue marble first, our bag with 3 red and 3 blue becomes a bag with 4 red and 2 blue, and vice versa. So when you draw a second marble, you either have a 1/6 or 2/6 chance of drawing a blue marble the second time depending on the colour of the first marble you drew.

And that concludes our lesson on the Types of Probabilities.


And without delay, I give your my scribe post for our May 23rd (Friday) class on Mutual Exclusive Events.

Slide One:
Contains a fox that is entirely unrelated to probability. However, I would like to dub him (or her) the Probability Fox, just for fun.

Slide Two:
A simple slide with a simple question. If you have 56 listed and 144 unlisted phone numbers, you have a total of 200 phone numbers. So if you want the probability of choosing a listed phone number from those 200 total numbers, you have a 56/200 chance, or a 28/100 or 7/25, or .28, or 28% chance.

Slide Three:
Okay, so we have a horse named Gallant Fox (makes total sense, right?) and another horse named Nashau. Gallant Fox runs a race and has a 2/5 chance of winning. Nashau runs an entirely separate race and has a 1/3 chance of winning. What is the probability that:

a) Both Nashau and Gallant Fox win their respective races. That's easy, we simply multiply their probabilities together and get 2/15. Remember earlier how our final probability was the product of all the probabilities of the steps before it? This is just like that, where our first step is Gallant Fox wins (2/5) or he loses (3/5), and then our second step is Nashau wins (1/3) or he loses (2/3). Since we want the path that has both Nashau and Gallant Fox winning, we want to multiply 1/3 by 2/5, which gives us our final probability for that outcome.

b) So this is just a in reverse, and instead of taking the winning path, we take the losing path. So 2/3 * 3/5 = 6/15.

c) Well, there are two ways we could solve this question. We could find the probabilities for all the routes where one of the horses wins their race and then add those together. But thats just long and tedious, and considering what we already know, we have a much simpler solution at our doorstep. The question asks us what the probability is that one of the horses wins, so as long as they both don't lose, we're good right? Wait, don't we already know the probability of them both losing? Well... if we know that, cant we just take all the possibilities and remove the ones we don't want to get the ones we do want? Something like this... 15/15 is our total possibilities, but 6/15 of them are ones that have both our horses losing their races. So we just subtract 6/15 from 15/15 and... our answer is 9/15.

Slide Four:
This slide was probably the most confusing question we had encountered in a while, and it took a while to figure out. But basically, it goes like this:

Chad wants to meet his girlfriend in either the Library or the Lounge. If he goes to the Lounge, he has a 1/3 chance of meeting her (apparently Chad has poor arranging skills as he still has no idea where they're meeting despite the fact that this has all been pre-arranged, but I digress), but if he goes to the Library, he has a 2/9 chance of meeting her.

a) What is the total probability he'll actually meet up with her, either in the Library or the Lounge. Because its an "or" probability, we add the probabilities together (the Library OR the Lounge), and get 5/9 chance they actually meet.

b) Now this is where it kinda gets confusing. The probability of them not meeting is 2/3 * 7/9? Wait, just were adding just a moment ago, why did we suddenly switch to multiplication? Now we're dealing with an "and" probability, where she is not in the Library AND not in the Lounge. Thus, we get our 14/27 chance they don't meet.

Slide Five:
This slide is a little demonstration on how Chad cannot go to the Lounge AND the Library. Thus, going to the Library or going to the Lounge are mutually exclusive events, which is what this lesson is all about.

Slides Six & Seven:
These slides give you the basic definition of Mututally Exclusive Events, and some examples. These are pretty self-explanatory.

Basically, if you have one event that makes another impossible, then they are mutually exclusive. You cannot turn left and right at the same time, you cannot be facing North and South at the same time. Facing North is mutually exclusive to facing South.

Slide Eight:
So we want the probability of drawing a King or a Spade in a single draw from a pack of 52 playing cards.

Our first event is to draw a King. Our second event is to draw a Spade. Drawing a king has 4/52 chances, and drawing a Spade has 13/52 chances. However, because of the one card that is both a King and a Spade, we must remove that card (1/52). So our formula:

Probability(Event A or Event B) = Probability(Event A) + Probability(Event B) - Probability(Event A and Event B)

Means this:

The probability of Event A or Event B is equal to Event A's probability plus Event B's probability subtract any Probabilities that fulfill both Event A and Event B.

And inputting our data, we get the Probability of drawing either a king or a spade as 16/52.

Slide Nine:
And finally, a couple practice questions. Drag'n Drop Baby!

a) Independent because the marble is returned, mutually exclusive because you cannot draw a red and blue marble at the same time.

b) Independent because there is only one step, not mutually exclusive because it is possible to draw a red king.

c) Dependent because once a president or treasurer is selected, the number of people in the selection pool changes and thus so do the probabilities for the next selection. Mutually exclusive because the president cannot be the treasurer at the same time (they are removed from the selection pool once they have been chosen).

d) Independent because there is only one step, mutually exclusive because there is no card that is a red king and black queen.

e) Independent because the second step does not change the probability of the second, mutually exclusive because its not possible to get an even and odd number at the same time.

And that concludes this super long double buttery extra flavoured scribe post, again posted in the middle of the night and written in the dark. I'll see you all in a few hours. Please feel free to point out any mistakes, I'm fairly sure I made a few.

And because he asked so nicely, my next scribe will not be Thi.



I'll just tell you guys in the morning since nobody will read this until then anyway. Good night, and have a pleasant tomorrow.

P.S. That Google translate thing works, atleast, I think so (I cannot read Japanese and verify this). However, from this I have learned that "Eleven" is "11" in Japanese.

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