why does math hate me? i have a problem with probability questions... i know that i can get a heads or tails - 1/2 but that's it. i have a few math books - i start reading about probability and i get all confused. could someone please explain it to me in very simple terms... i know that (and) means muliply and (or) means addition - but when the problem is written i just don't get how to apply it. any help greatly appreciated!

Jello, imagine a regular die (dice is a plural). Better yet, get hold of a pair of dice. (Go ahead, I'll wait.) The die has six sides, and each side has a different number of spots on it (or a different number), from 1 to 6. Six sides means there are six different possible outcomes from a single roll (well, assuming the die is fair). Now how many 2's are on the die? One, right? Therefore the probability (odds) of rolling a 2 are one in six, otherwise known as 1/6. And the odds of rolling a 1 would be... Now consider this question: What are the odds, on any given roll, of rolling a number that's less than 3? Well, we still have the same six possible outcomes. But which of those possible outcomes gives numbers less than 3? Two numbers do: we win if we roll a 1, and we win if we roll a 2 - that is, we win if we roll (1 OR 2). At this point, you may certainly add 1/6 (for the odds of rolling the 1) + 1/6 (for the odds of rolling the 2), which gives 2/6, which reduces to 1/3, and it might be wise to put it that way in a CSET constructed response. It's also perfectly valid to notice that a total of 2 outcomes out of the possible 6 will do the job, given 2/6 which reduces to 1/3. You could work out the odds of rolling an even number, or a number between 2 and 4, exactly the same way. You could even list all the outcomes - 1, 2, 3, 4, 5, 6 - and circle the outcomes that are less than 3. So OR works with a single roll, or spin of a spinner, or digit in the lottery, or whatever. In contrast, AND doesn't, because it's just not possible with a die to have a throw that comes up 1 AND 2 at the same time. Where AND comes in is where we have more than one event - that's well illustrated by two dice thrown at the same time. What are the odds of rolling a pair of 2's? We could do this by listing outcomes: 1+1, 1+2, 1+3, 1+4, 1+5, 1+6, 2+1, 2+2... and so on. How many possible outcomes are there with two dice? Count 'em, and you get 36. And if you keep listing the pairs as I've done, pretty soon you'll begin to realize WHY there are 36: because each of the six numbers on the first die can pair with any of the six numbers on the second die, and that's 6x6. And how many of those outcomes give us a pair of 2's? Exactly one: 2+2, which gives us 1/36. Or you can do this mathematically, multiplying the odds for a 2 on die A times the odds for a 2 on die B, which is 1/6 x 1/6, which still gives 1/36. If this doesn't help, for heaven's sake keep asking!

Very well explained! Now, in situations where dice are not involved. Let's say you are going on a two-week trip and you packed 3 shirts (white, green, red) and 3 pairs of pants (tan, blue, plaid). Do you have enough outfits to wear a new outfit for every day of the trip? First thing you need to know is how many different combinations of shirts and pants you can arrange. This is the denominator of the P fraction (how many results in all). Make a little graph or tree diagram. Start with one color shirt and write down all the combinations. White shirt - tan pants White shirt - blue pants White shirt - plaid pants Now do the same for the other two shirts. Green shirt - tan pants Green shirt - blue pants Green shirt - plaid pants Red shirt - tan pants Red shirt - blue pants Red shirt - plaid pants How many total possible results (outfits) are there? 9 You don't have enough for a two-week trip. What is the probability that you will pick out a particular outfit with your eyes closed? ie, what is the P(red shirt/tan pants)? Look at your chart. How many times does the combination red shirt/tan pants occur? Once. The P(red shirt/tan pants) is 1/9. What is the probability that you will pick out the red shirt/plaid pants OR white shirt/plaid pants? P(red shirt/plaid pants) + P(white shirt/plaid pants) = 1/9 + 1/9 = 2/9 In any probability question, the first thing to do is to identify how many different results there can be. Make a list. Then circle how many times the desired result is listed. Here's another example. You have a spinner with 8 sections: 2 are red, 3 are blue, 1 is green, 2 are yellow. What is the P that you will spin a yellow? List all the possible results of one spin: red, red, blue, blue, blue, green, yellow, yellow. The P is the number of desired results divided by the total number of occurrences. There are 2 yellows out of 8 possible results. P(yellow) is 2/8. Reduce the fraction. P(yellow) is 1/4. P(yellow or green) is 2/8 + 1/8 = 3/8. P(yellow AND green) (on two spins) is 2/8 X 1/8 = 2/64 = 1/32. Once you know how to list out all the possible results you can always find the probability.

thanks teachergroupie! you really clarified that a lot... i had to read the second AND part a few times - but now i get it. i just did some problems below to double check and practice a constructed response.. what are the odds for rolling an even number on the a fair die? there are three even numbers on one die: 2-4-6. we win if we roll a 2 OR 4 OR 6, so you could add the following to get the answer.. 1/6+1/6+1/6= 3/6 =1/3 for a number between 2 and 4: there is only one number between 2 and 4 so the odds would be 1/6 for the AND problems - i think i will use "you can do this mathematically, multiplying the odds for a 2 on die A times the odds for a 2 on die B, which is 1/6 x 1/6, which still gives 1/36" this makes a lot of sense.

thank you upsadaisy - that cleared things up a lot. since there is a time restraint for each question on the test.. is there a faster way to figure out the combinations of the outfits you stated above - instead of making a tree? thanks again!

i guess to get the total outcomes 9 combos would be easy... but if they asked about the other combos... red shirt/blue pants or white shirt/green pants for example - with a tree it would be easier to visualize. or is there also a way to figure that out without a tree? thanks

If you know how many times that would occur you can just put that number over 9. If there was only one red shirt and one blue pants, then it could only happen once - so P is 1/9. However, if you had 2 red shirts and one pair of blue pants, it could result twice, so the probability would be 2/9. Now, try this - what if you had a color spinner (3 red, 2 yellow, 1 blue) and a regular die. What would be the P of getting a red and the number 6?

Um, 3/6 reduces to 1/2, not 1/3... but yes, odds of rolling an even number on a fair 6-sided die are 1/2. What the odds are of rolling "a number between 2 and 4" depends on how you define "between 2 and 4". If it's "between 2 and 4 INCLUSIVE", that means 2, 3, 4, and the odds would be 1/2; if it's EXCLUSIVE, that means just 3, and then you're right with your 1/6.

The outfit problem is a problem in combinations. It's probably a little tidier to consider probabilities and combinations separately. The probability question asks "what is the probability" or "what are the odds" or even - very occasionally - "how likely?" The combination question will always ask "how many different combinations" - or outfits or meals or whatever is being referred to. Suppose we're making up bagged lunches and we've got peanut butter sandwiches, bean burritos, and tofu wraps; carrot sticks, celery sticks, zucchini sticks, and radishes; bananas, apples, mandarin oranges, and bagged grapes; and milk, Gatorade, and water. How many lunches can we make? (Actually, you COULD work out the odds, if the lunches are all made up and are in unmarked bags, that a kid will grab a lunch with a wrap, radishes, an apple, and water... go ahead and give that a whirl.)

Blush. I should have written my combination problem like this: Suppose we have a peanut butter sandwich, a bean burrito, and a tofu wrap; a bag of carrot sticks, a bag of celery sticks, a bag of zucchini sticks, and a bag of radishes; a banana, an apple, a mandarine orange, and a bag of grapes; and a container of milk, a container of Gatorade, and a container of water. How many different lunches could we make? I screwed up in part by asking how many lunches we CAN make, when I meant to ask how many lunches are POSSIBLE. Obviously, if every lunch has to have items from each category in it, there can't be more than three lunches. To figure out how many lunches are POSSIBLE without using the tree diagram, multiply together the number of items in each category. As the numbers of categories and/or members of each category go up, the numbers of possibilities go up rather radically fast. To calculate the odds on (=probability of) a lunch having a wrap, radishes, apple, and water, find the fractions that represent the odds in each category - you should end up with four fractions - and multiply them together.

TeacherGroupie - once you make a tree for the possible combinations you should find that the wrap, radish, apple, water combination occurs just once. Therefore, the probability would be 1/180. (You might have been referring to the P of picking those items in four separate events ???) Probability isn't as easy as it seems at first.

For the most part, CSET questions involve EITHER combination OR probability but not both together. That simplifies things considerably. My point was that you don't HAVE to make a tree for the possible combinations: you can calculate them. 3x4x4x3 = 144 distinct possibilities. Neither do you actually HAVE to make a tree to find out the probability of the wrap/radish/apple/water combo: The probability of wrap is 1/3, as is the probability of water; the probability of radishes is 1/4, as is the probability of apple. 1/3 x 1/4 x 1/4 x 1/3 = 1/144. To calculate the probability of non-sandwich, non-root vegetable, tree fruit, and milk, first realize that non-sandwich means burrito OR wrap, or 2/3; non-root vegetable means zucchini OR celery, or 2/4 = 1/2; tree fruit means banana OR orange OR apple, or 3/4; milk is still 1/3. Multiply 2/3 x 1/2 x 3/4 x 1/3, and feel free to cross-cancel... the result is 1/12. And this is faster than doing a tree, no? Probability on the level demanded by CSET is actually not all that complicated - the trick lies in picking the terms apart and then putting them back together. Of course, that's true of all math.

i am relieved that the cset will not be mixing the combination with probability! this further explanation really helps... so i can multiply the groups together to figure out the combos... and with probability i can take each group separately and than put them together. i had a problem last night in a prep book - it was a combo lock with three slots has ten possible numbers from 0-9, and they wanted to know how many combos there are for the lock... i drew a blank.. now i see there are three groups (the answer said someting about them being independent) so i can multiply the three groups together to get the different combinations - 10x10x10 = 1000 so this is an example of an independent event... what would i do with a dependent event... off to look for answer... and to have a sandwich wrap, some radishes, grapes and water

"Independent" here means that getting (say) 8 in the first slot doesn't make 8 unavailable for the other two slots. Now suppose instead that you've got a box with 10 marbles in it, the marbles numbered 0 through 9, and you're going to pull out three marbles without putting any back. What are the odds of pulling out the 3, then the 4, then the 2? Well, the odds of pulling out the 3 are 1/10. The odds of then pulling out the 4 are 1/9, because there are only 9 marbles left; then the odds of pulling out the 2 are 1/8, because there are only 8 marbles left. Multiply together, and you get odds of 1/720 - better, I might add, than the odds on the average slot machine, but still nothing to bet the mortgage on. I trust you enjoyed the wrap, radishes, and grapes, but how about some nice wine instead of the water?

a nice Chianti sounds delicous! do you know the properties in wine and dark chocolate (70+ percent cacao) that make them good for you? that makes sense about the odds and the marbles - i would probably mess that up because i would forget to subtract from total amount once i took out the marble. This is my biggest problem with math - the small details and sometimes writing a number wrong... today I was checking out a formula that really helped me with some of the mixed combination problems: (don't start yawning yet!) nCk= nPk/k! for example Anna has 4 pictures, how many different pairs can she make? nCk=4P2/2! = 4x3(the 4 on right of P is mulitiplied going down from 4 to number 3 - these two numbers 4x3 are the P2 part of problem) 12/2! (2! means 2x1 because the factorial means going down from 2 to 1, for example 6! =6x5x4x3x2x1) 12/2=6 looks like the problems I have seen from cset site and cliffs book (which seems more aligned with actual test) the probability questions are a little easier...

Jello, you're playing my song... just let me grab a corkscrew, and... The stuff in wine and chocolate that's good for us? Phytochemicals, a.k.a. flavonoids - all in the antioxidant category. They're present in milk chocolate as well, just accompanied by more sugar and fat. (Ask me if I care. Education runs on chocolate!) Probability questions on the test aren't really that complicated once you figure out what's going on. If you search the A to Z site, you might find my riff on four-faced dice (they look like little pyramids with triangular bases) - think about the odds of rolling a 3 on one of those as opposed to a regular six-sided die, or for that matter the odds of rolling a 3 on a dodecahedron (twelve-faced) die: each face is a pentagon) or an octahedron (eight-faced die, each face a triangle). I think you're going to be fine on probability and combinations now. Did I remember to recommend Usborne's nifty DICTIONARY OF MATH?