Probability: More Practice

Instead of starting a new unit, we worked on Probability. Mostly we went over the practice sheet we were given on Thursday. The questions we reviewed were one, two, six and nine.


Question One:

• The probability of a randomly chosen car being defective is 1/3. Four cars are chosen randomly in order. Given that at least two cars are defective, what is the probability that the first car is defective?

The first thing we did, was figure out what the question was asking.
P(1st is defective I at least two is defective)

Next we found the sample space. It's always easier if you think of it as a word, and then asking yourself how many times the letters in that word can be rearranged, like so:
DDGG -- meaning two of the cars are defective
DDDG -- three cars are defective
DDDD -- all four cars are defective
Then you can find the probability that the cars will be chosen that way, and that becomes your sample space, or denominator.

Next we looked at all the possibilities that the first car will be defective using slots, which, if you remember, was from combinatorics.

Put that numerator on top of the denominator that we found earlier, and you have your answer.



Question Two:

• Two jars contain red and green marbles. Jar I contains 3 red and 2 green marbles. Jar II contains 4 red and 3 green marbles. A jar is picked at random and two marbles are picked out of that jar in order. If it is known that the first marble is red, what is the probability of the second marble being red?

You can start this problem by drawing a tree, and writing down the probability that each possibility will occur - if that makes sense. For example, if you look at the picture above, you'll see that the first branch is 'Jar I' followed by 'Jar II', the probability of choosing either jar is one half. The next branches are 'red' and 'green', and the probability of choosing red in 'Jar I' is 3/5, while the probability of choosing green is 2/5. So on and so forth. Keep in mind that the marbles are not replaced after you take it out, so the sample space is different everytime you draw.

In this question, what we're looking for is the probability of getting a red after drawing a red marble first. That becomes the numerator, and the denominator consists of all the possibilities, as shown above. So you plug in the proper numbers in their proper places, and you have that question solved.



Question Six:

• Susan sees her friend, Tim, at his locker with a worried look on his face. She asks, "What's wrong?" Tim has to open his locker and change clothes within the next five minutes. However, he has forgotten the combination to his new lock. He knows that the lock requires three different numbers. He also remembers that all of the numbers are odd, and all of the numbers are divisible by seven. It takes 10 seconds to dial a locker combination and 1.5 minutes to change clothes. Is Tim likely to be ready for gym class on time?

I think you'll understand how we solved this problem just by looking at the above picture. The '1 min 30 sec', written in pink, is how long it takes Tim to change.



Question Nine:

• Three identical boxes each contain two drawers. In one box, each drawer contains a gold coin. In another box, each drawer contains a silver coin. The remaining box has a silver coin in one drawer and a gold coin in the other. One drawer is opened and a gold coin is found. What is the probability that the other drawer in that box also contains a gold coin.
• Michael claims that the probability is 1/3. Jessica claims it is 1/2. Raymond says the probability is 2/3. Explain how each person may have arrived at their answer. Who is correct? Justify your answer.

The probability of choosing the right box is 1/3, and the probability of choosing either a gold or silver coin is 1/2, since there's two drawers. Obviously if you get the box that only have two gold coins, you're not going to get a silver coin. But I'm sure we're all smart enough to realize that.

The probability of getting a second gold coin is shown by:
[P(IG)/(P(IIIG) + P(IG) + P(IG))] + [P(IG)/(P(IIIG) + P(IG) + P(IG))]

If we plug in the correct numbers, we have:
[(1/6)/((1/6) + (1/6) + (1/6))] + [(1/6)/((1/6) + (1/6) + (1/6))]
[(1/6)/(1/2)] + [(1/6)/(1/2)]
(1/3) + (1/3)
2/3

Therefore, Raymond is correct. Michael could've gotten his answer by thinking that there's only one box that contains two gold coins, and the probability of choosing that box is 1/3. Jessica could've simply thought, well there's two drawers, so there's a 1/2 chance of getting a gold coin.



So that's the end of that. There's also questions in the slides (9-18) that we could do for practice. In the afternoon class we had a pre-test that we're going to go over tomorrow. There isn't going to be a test for this unit because we don't have time. Soo.. we're starting a new unit next week.

And that's that. The next scribe will be Joseph because he asked me yesterday. Whoot. Good luck to everyone doing the English Provincial Exams tomorrow! Remember to pace yourself and.. uhh, I forgot what else Mr. V and Mrs. Strecker said.. but.. yes!

Today's Slides: May 30

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Workshop Period

Ohhh my goodness. I nearly forgot I was scribe today because of the basketball game XD. (Go lakers!)

So the title above describes todays class. Mr. K was away again but no worries, he'll be back tomorrow...i think. Anyways, we were given a worksheet that we worked on in our little groups for the whole class. I thought the sheet had some challenging question like 1 and 2. I hope we go over those tomorrow. Yeeaah it was a pretty uneventful period other than that.

So I'll get to the point. Next scriibe is NELSA. "Beautiful" right Nelsa?

Today's Slides: May 27

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Combinations of Permutations and Probability and Combinations O_o

Why hello there!
Benofschool here. I never scribed for a while, I was beginning to think that you guys forgot about me =(. Well anyways time for a scribe. Mr.K was later than usual today but he wasn't late for school which was good. Today's class was a workshop. We were broke up into groups like every other workshop class. Unfortunately the slides aren't up. Questions were put up and we were off...

Question 1

We had to find the probability of getting an ace or a diamond from a deck of cards. Fairly simple. Francis went up to answer it. To get the answer we just add the probabilities of getting an Ace and a Diamond and subtract the probability of getting the Ace of Diamonds. We add because it is an "or" question and we subtract that single card because it was counted twice when we calculated the probability of getting an Ace or Diamond. So the answer was 4/13.

Question 2

The second question involved a venn-diagram and venn skills from grade 11 logic in pre-calculus. The question was what is the probability of picking a person who doesn't like Dr. Pepper or the diet version if 7 people liked Dr.Pepper, 11 people liked the diet version, and 3 people liked both. So first we have to create a two circle venn-diagram. Rule is start with the inside and work outwards. So we put in the middle which means that 3 people prefer both drinks. Then lets work with on side. But remember from grade 11 we have to subtract the number of objects in the center from the separate values. If we look at the first question, we didn't count the Ace of Diamonds because the probability of getting an Ace or a Diamond is not mutually exclusive as well as the Dr. Pepper situation. If one thing occurs, the other probability can still happen. So that means we have to subtract the number of people that liked both drinks from the separate preferences. We get the sum of those numbers (4 +3+8=15) and subtract from the total number of students in the class (25-15=10). That will be the numerator in the probability and the sample space would be the total number of students. So the answer is 2/5.

Question 3

The 3rd question is about picking dresses. There are 15 dresses: 6 green, 5 blue and 4 yellow dresses. We wanted to know what is the probability of getting exactly 2 green dresses from picking 6. This question involves both combinations and probability. So 6 choose 2 because they are indistinguishable. multiplied by 9 choose 4 for the remainder of the dresses divided by 15 choose 6 which is the ways that we can choose 6 dresses out of 15. So we get an answer of 37.8%.

Question 4

Another combination question. A couple has 4 children that are about to be born. What is the probability of getting at least 2 girls. So what we do is find the probability of getting 2 girls first then 3 girls and finally all girls. Then we just find the sum of those values and we get the answer. The answer is:

That were all of the questions we did in today's class. Remember DEV due dates are arriving in a future now. Check the calendar on the right side bar to check for your due dates. Tomorrow's double will be used as a DEV work day. So bring the stuff you want to bring to work on your DEVs. That is all I have to say about today's class. The next scribe will be kristina. Good Night and see you all tomorrow!!!

why DOCTORS are so DARN CAGEY!

and no ...
- being cagey does NOT mean you're inside a cage alot ...

So I was chosen for being the scribe for today. Let me tell you guys something. REVENGE GETS YOU NOWHERE. That's why I'm always going to steal Paul's seat from now on. Thanks for choosing me as scribe Paul.



[ SECTION A ]
- SLIDE 2, 3
- Identifying events / dice diagram


[ SECTION B ]
- SLIDE 5, 4
- Probabilities involving "AND" and "OR"
- Testing for independence


[ SECTION C ]
- SLIDE 6, 7, 8, 9
- A test for cancer


[ SECTION D ]
- SLIDE 10
- A test for industrial disease


[ SECTION A ]
- identifying events / dice diagram


SLIDE 2 - IDENTIFYING EVENTS

- and so we started the class finishing off what we had last done in class on Friday.
DRAG N' DROP BABY!
- a list of events which the class had to determine whether it was dependent or independent. Mutually exclusive or not mutually exclusive. I'll take two examples.

b) one card - a red card or a king - is randomly drawn from a deck of cards.
- because it is only ONE event we know that it must be independent
- a red card can also be a king. Therefore it is not mutually exclusive

c) A class president and a class treasurer are randomly selected from a group of 16 students.
- because we can only have ONE president and ONE class treasurer, the event must be dependent. One student cannot be the president and class treasurer. Well in this case it can't. Therefore it is mutually exclusive

SLIDE 3 - THE DICE DIAGRAM

FORGIVE ME IF I AM DOING AN ILLEGAL ACT ON THE INTERNET.

I had actually found this diagram on another blog of Mr. K's class.

This was done to answer the question:
e) Rolling two dice and getting an even sum or a double
- It is independent because the first roll does not affect the chances of the probability for the next roll.
- As you can see on the diagram, you are able to roll a double AND get an even sum. There fore it is not mutually exclusive.


[ SECTION B ]
- Probabilities involving "AND" or "OR"
- Testing for independence


SLIDE 5 - PROBABILITIES INVOLVING "AND" OR "OR"
AKA "THE ADDITION RULE"

here is the slide ...


SLIDE 4 - TESTING FOR INDEPENDENCE



We compared the result of the probability of getting the flu shot and getting the flu. (0.10)
To the result of the probability of the seniors getting the flu (0.15)

There fore the event is dependent if you get the flu shot or not.


[ SECTION C ]
- A test for cancer


SLIDE 6, 7, 8, 9 - A TEST FOR CANCER


- The given question, information and possibilities.


0.5% of 1,000,000 = 1,000,000 x 0.005 = 5000
To find who does not have cancer you would subtract 5000 from 1,000,000
-
98% of the time, the test will be positive.
The amount of people who have cancer and will test positive would be ..
5000 x .98 = 4900
The amount of people who have cancer and will test negative would be ..
5000 x .02 = 100
982% of the time, the test will be negative.
The amount of people who don't have cancer and will test positive would be ..
995,000 x .02 = 19,900
The amount of people who don't have cancer and will test negative would be ..
995,000 x .98 = 975100
-
Adding it up the amount of people who tested positive for cancer
4,900 + 19,900 = 24,800
When only 4,900 actually do have cancer.

The probability that a person who actually does have cancer and tests positive is the amount of people who do have cancer divided by the total amount.

4,900 / 24,800 = 19.75%

Here are the tree diagrams to see how this all works out.




[ SECTION D ]
- A test for industrial disease


SLIDE 610 - A TEST FOR INDUSTRIAL DISEASE



The probability of a person who tested positive and actually does have the industrial disease is much similar to what was done in SECTION C.

The probability of people with industrial disease and tested positive.
DIVIDED BY:
The probability of people with industrial disease and tested positive,
ADDED WITH:
The probability of people without industrial disease and tested positive.

^ does that make sense?!

In other words, and in this case ..

0.0099 / 0.0099 + 0.0099 = 0.5%



NO WONDER DOCTORS ARE SO DAMN CAGEY! ...
... BECAUSE THEY'RE ALWAYS INSIDE CAGES! WOW!
because there is still a great chance that even if you do NOT have cancer, the result may still test POSITIVE

SO YOU MIGHT HAVE CANCER, MUAHAHAHAHA
- not intentionally trying to scare any of you, or myself.
- not intentionally trying to offend anyone who is reading this.
- It's just a joke! Humor in OUR math class is NO laugh matter.

That's it, That's all ladies and 'gents! Have yourself a great day. Go out there and commit random acts of kindness! Cheers!

The next scribe will be ...benofschool

Today's Slides: May 26

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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...

TYPES OF PROBABILITY: 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.

EXCLUSIVITY IN PROBABILITIES: MAY 23

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.

No, the next scribe will be SOMEONE WHO HASNT BLOGGED YET, WHY ISNT THE SCRIBE LIST UP TO DATE WHAAAAAARGHADSADJKSAHDJKAHSJDL.

>:(

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.

Today's Slides: May 23

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Today's Slides: May 22

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An Introduction to Probability

Alright young fellers, american idol is almost on so I'll try to finish this before it starts but if I don't..I'm sure I can do both! (:

Slide One
The first thing Mr. K asked us today was, "What is the probability of the sky falling?" (chicken little yay!) Each of us were given about a few minutes to write down our answer and we all shared our answers to each other. *refer to the first slide* Most of us came up with different numbers but some of us had the same answer as others. Most of us just chose a random number, like me, but some of us also took some time to why they choose that number. After this dicussion, he then asked us what the numerator and the denominator represented. The numerator is the number of chances of the sky actually falling while the denominator is the set of all possible outcomes.

Slide Two
The next slide explains the defintion of an inderterminate form. So if we take any number such as 6 and multiply it by one then divide by one your answer would be 6. And if you take the 6 again and multiply it by 1/10 then divide by 1/10, you still get the answer 6. Continuing the pattern, you'd eventually reach zero. So you would multiply 6 by zero then divide by zero. Therfore it is in an inderterminate form. If you recall on slide one, 0/0 and 0/∞ are examples of indeterminates as well as ∞/∞ and ∞/0. 1/0 is not in an indeterminate form, it is however, an undefined value.

Slide Three
The next slide includes various vocabulary words which we should know for this unit as well as different examples to explain the meaning of each words.

Slide Four
Calculating the Probability of Event A
This slide includes the formula P(A) = n(A)/n
P(A) - Denotes the probablity of A
n(A) - number of favourable outcomes (number of accourences of A)
n - sample space (the total number of possible outcomes
Probability can not only be expressed as a fraction but also as a ratio, decimal and a percent. However we will not be looking at it as a ratio in this unit.
IMPORTANT TO REMEMBER: Probability of any event is always a number between 0 or 1. This means that if your answer is bigger than one or it is a negative, then your answer must be wrong.

Slide Five
The fifth slide is also a continuation of the different vocabulary. In this specific slide we dicussed mainly about the complimentary event. The example he asked was, if there was a sudden death single tennis tournament and there were 54 players, how many games would be played if there could only be one winner? The easiest way of coming up with the answer is asking yourself, "how many losers would there be" The answer is 53 because only one player could win. Another example is using a coin or a die. Basically, you're looking for the opposite. So if P(E) = a then P(E¹) = 1-a.

Slide Six
In the next slide, we had to determine the sample space when a fair die is rolled once. Since it is possible to roll either 1, 2, 3, 4, 5 or 6, the answer to the question is {1, 2, 3, 4, 5, 6} *note: if you are asked to determine the sample space you must use brace brackets { }*
The next question was to determine the sample space for rolling a six sided die and flipping a coin. You could either use a tree diagram or a chart to find the answer. In this case we first used the tree diagram to find the answer. The first two branches consited of Heads and Tails and each had 6 branches to indicate the numbers on the die. Each number could either get H1 to H6 or T1 to T6 so the answer was 12.

Slide 7
We then had to determine the probability of rolling a 2 when rolling a fair die. The answer is 1/6 because there's only once chance of getting a 2 out of 6 different outcomes. We write this as P(2) = 1/6
In the next question it asked to determine the probability of getting a head and an odd number when rolling a die and flipping a coin. In this case we used the previous chart we made on slide 6 to find the answer. We then found that there are 3 possible chances out of 12 possible outcomes which also redues to 1/4. *note: remember to always reduce your fractions.

I believe I covered most of what we learned and talked about in class, so that means..I'm finished! (: This scribe actually took longer than I clearly expected considering I was running back and forth to my room and my sister's to watch American Idol. But enough of my life..toodles everyone, American Idol isn't over yet!

BTW..Homework is on slides 8-10 and don't forget to study for tomorrow's test! Goodluck and Ciao! OH and.. Next scribe is... "what it do PAUL wall" (:

Today's Slides: May 21

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