The basics (if you're comfy with the basics, skip to the big reveal.)
The probability of an event is equal to the number of ways that event can occur divided by the total number of possible outcomes.
So the probability that you will be chosen at random to represent your 30-person class in the hot dog eating contest is 1/30. Likewise, the probability that your frenemy Ashley will be chosen for the contest is 1/30.
What if I asked you about the probability that EITHER you OR Ashley would be chosen? Well, now the event we're concerned with happens in 2 of the 30 possible outcomes. It would be satisfied if you were picked, or if Ashley was. Therefore, the probability is 2/30, which simplifies to 1/15. When there are multiple, mutually exclusive ways an event can occur, ADD the probabilities of each way to get your overall probability. Again, this only works if you're talking about events that are mutually exclusive. In other words, it only works if both events can't happen at the same time. Example: if you buy lottery tickets, since only one ticket can win, every ticket adds to your probability of winning.
This is important, so make sure you're solid. Hover your mouse over the following examples to ensure this won't leak out of your brain.
- What's the probability of rolling an even number on a standard 6-sided die?
- What's the probability of rolling a number less than 6 on a standard 6-sided die?
- What's the probability of rolling a prime number on a standard 6-sided die?
- What's the probability of flipping a coin and getting either heads or tails?
- What's the probability of flipping two coins and having one or the other come up heads?
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| Yo dawg I heard you like dice. |
This rule only holds if the events are independent; whether one occurs has no effect on whether the other occurs. When one event's occurrence effects the probability of the next event's occurrence (we say the events are dependent), the rules change a little bit: Say there was a rule that the winner of the hot dog eating contest was ineligible to be chosen for the vomit cleaning contest. In that case, the probability of Ashley being chosen for the second contest (given that you were chosen for the first contest and thus ineligible) becomes 1/29 (we don't count you as a possible outcome anymore). So if the hot dog contest winner couldn't be chosen for the vomit contest, the probability that you'd be chosen for the first and Ashley would be chosen for the second would get very slightly higher: (1/30)(1/29) = 1/870.
We still good? Prove it (again, mouse over the questions to get the answers):
- What is the probability of flipping 4 coins and having them all come up heads?
- John and Sam are both choosing randomly from 5 types of candy: types A, B, C, D, and E. What is the probability of them both choosing candy type A?
- Names are being picked out of a hat. If there are 8 different names in the hat, what is the probability that Sven is picked first and that Gretchen is picked second?
Probability of X or Y = (Probability of X)+(Probability of Y)*
*(as long as X and Y are mutually exclusive)
*(as long as X and Y are mutually exclusive)
Probability of X and Y = (Probability of X)×(Probability of Y)
Ready for the big reveal?
It's nice to know all that stuff above, but you don't always need it! Most of the time, it's sufficient just to list all the possible outcomes, and count the ones that match your conditions, especially on the hardest questions! Seriously, the SAT always plays with small numbers (remember: calculator optional), so it's never too onerous just to put pencil to paper and start listing. Example:Mathematically, there are 3 ways to get 2 heads and one tails: HHT, HTH, THH. There are 2×2×2=8 total possible outcomes, so the answer is (B) 3/8. Note, though, that it requires some thought to come up with the possible outcomes that satisfy our condition. In the time it took you to think about possible HHT combinations, you also could have just listed the 8 possible outcomes, and counted.
- What is the probability of flipping 3 coins and having 2 of them come up heads and 1 come up tails?
(A) 1/3
(B) 3/8
(C) 1/2
(D) 5/8
(E) 7/9
Possibilities (satisfactory outcomes bolded)
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Let's look at one more (slightly tougher) question:
- Phil is holding 4 cards in his hand: 8 of clubs, 5 of hearts, king of hearts, and ace of diamonds. If he places them on a table in random order, what is the probability that the first and last cards will both be hearts?
(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/6
(E) 1/8
Step 1: Assign numbers to the cards (give 1 and 4 to the hearts for simplicity's sake).
1 - 5 of hearts
2 - 8 of clubs
3 - ace of diamonds
4 - king of hearts
Step 2: List all the possibilities that start with "1" (ones with hearts on the ends bolded).
1234
1243
1324
1342
1423
1432
Step 3: Either list all the possibilities that start with "2," or recognize that all the possibilities that begin with "2" or "3" cannot possibly satisfy our condition of having hearts on both ends because "2" and "3" are not hearts, and skip right to listing all the possibilities that begin with 4 (again, bolding the ones with hearts on the ends).
4123
4132
4213
4231
4312
4321
Step 4: Count the successful outcomes (there are 4 of them), and count the total possible outcomes (we didn't list the ones that began with 2 or 3, but there are 6 each of them, just like there are 6 that begin with 1 and 6 that begin with 4. Total: 24 possible outcomes.
Answer: The probability of getting hearts at the beginning and end is 4/24, or 1/6. That's choice (D). Full possibilities list:
1234 2134 3124 4123
1243 2143 3142 4132
1324 2314 3214 4213
1342 2341 3241 4231
1423 2413 3412 4312
1432 2431 3421 4321
Again, this is really the way I do these questions when I take the test. I respect it if you want to solve them the math way every time, but I caution you that such a dogmatic adherence to math on a test that is NOT A MATH TEST increases your likelihood of making a mistake under pressure. Your call.
In which Corey wonders why we don't just flip coins
- A deadly snake pit trap laid by a diabolical supervillian contains green and red snakes in equal number and of equal likelihood to attack. If Corey falls into the trap, what is the probability that the first 4 snake bites he receives are all from green snakes?
(A) 1/2
(B) 1/4
(C) 1/8
(D) 1/16
(E) 1/32
Note: Figure not drawn to scale.
- In the figure above, the circle is tangent to both the top and the bottom of the rectangle. If the area of the rectangle is 10, what is the probability (rounded to the nearest hundredth) that a point chosen at random inside rectangle also falls inside the circle?
(A) 0.63
(B) 0.59
(C) 0.44
(D) 0.31
(E) 0.28
- Regina rolls two standard 6-sided dice and is astonished to discover that both individual dice show prime numbers, and their sum is also a prime number. What is the probability of this outcome?
(A) 1/18
(B) 1/9
(C) 1/8
(D) 1/6
(E) 1/4
Also try this recent Weekend Challenge question involving probability (post contains multiple explanations).
Answers:
17. (D)
19. (D)
20. (B)
I will read this CAREFULLY later.....but in the meantime, you should check out my friend Catherine's blog. She is asking about probability (I left your post in the comments). http://kitchentablemath.blogspot.com/2011/04/help-desk-probability.html
ReplyDeleteThanks Debbie, I just did check it out. Looks like a few commenters have gotten it right the same way that I would, which is just to list the possibilities and count. It's a very similar problem to the #20 I posted in the drill above -- there's no neat and tidy way to just write an equation and solve, because the event you're looking for is itself defined by some mathematical relationship. Crazy stuff :)
ReplyDeleteWell I'm glad to know that the way I would have done it (i.e. what seems like the crazy long way) seems to be the "right" way to have done it.
ReplyDeleteMy question is this: How can you do that quickly enough on the SATs when you have about a minute per problem?
A little practice goes a long way, there. In the kitchen table math post, there are only 8 possibilities, so it really should take 20 seconds max to list them all IF YOU'RE COMFORTABLE with probabilities. That's actually what this whole post is about. Probability is one of the trickiest things the SAT will throw at you (thankfully, it's also rare), but almost all probability calculations can be sidestepped quickly and efficiently by listing possibilities and counting them. The trick is for that to be your first instinct! Lots of students wrestle with calculations for a minute first, before they even think to try listing possibilities. If you just start writing possibilities as soon as you see a question like this, you can actually get through it very quickly.
ReplyDeleteOk. THANK U! I have bookmarked the post to come back to later this evening. Off to pick up my daughter now and then running around. Actually, one of my errands is to stop off at Kumon and check out the vibe there. I had a thought last night which is that maybe I'll do Kumon over the summer with my kids. What do you think of Kumon?
ReplyDeleteIf only I could spend all day doing SAT problems (what I actually want to do). Morning was blown with hours and hours of paperwork.
Ah, paperwork. What a hateful task. :) I have similar wishes: if only I could spend all day WRITING SAT problems.
ReplyDeleteI don't have any strong opinions about Kumon either way, but I know some people swear by it.
Are you an official SAT question writer?
ReplyDeleteNo no no...I just impersonate one fairly well (I hope). I don't imagine they'd be happy if their official writers had blogs like this one.
ReplyDeletemate is there something wrong with problem number 19 ?? How can area of rectangle be 10 ?? and width = 5 ? that would mean
ReplyDelete2(L + W) = 10;2(5 + w) = 10;5 + w = 5;W = 0? So how is that possible ??
In number 17 too I didn't know how to solve it,because its not written how much snakes are actually inside each pit is there something wrong I interpreted or read wrong ?
ReplyDeleteYou're thinking of perimeter. Area = length TIMES width.
ReplyDeleteYou don't need to know how many snakes are there. A snake can bit more than once. So even if it's just 1 green snake and 1 red snake, the solution is the same. If you don't like snakes, this is basically just a coin flip problem. What is the likelihood of flipping a coin and getting heads 4 times in a row?
ReplyDeleteHow did you do number 20? I keep getting 1/18 D:
ReplyDeleteThis problem is in my book, too, so I'm going to paste part of the handwritten explanation I did there. I'm betting the problem is that you're not counting both 2,5 and 5,2. It's confusing, but you have to count them twice since they're they're twice as likely as a double (like 2,2) to come up.
ReplyDeleteThere are five boys and five girls from which we assemble a three-member
ReplyDeletecommittee. What is the probability that the committee consists of two
boys and one girl?
First, figure out the total number of possible committees. If order mattered, we'd to 10*9*8 and call it a day, but order doesn't matter in a committee; Suzy, John, and Sam is the same grouping as John, Suzy, and Sam. So we count the number of possibilities either using combinations, or by listing. In this case, because listing would take a while, we'll do it the mathy way: 10C3 = 10!/(3!7!) = (10*9*8)/(3*2*1) = 120.
ReplyDeleteNow figure out how many ways you could get 2 boys and one girl on a committee. For this, I'd list. Say the boys are A, B, C, D, E, and the girls are 1, 2, 3, 4, 5. Here are all your possible combos:
AB1
AB2
AB3
AB4
AB5
AC1...AC5
AD1...AD5
AE1...AE5
BC1...BC5
BD1...BD5
BE1...BE5
CD1...CD5
CE1...CE5
DE1...DE5
That's 50 different possible groupings that have 2 boys and one girl. Note that once you see how things are shaking out, you can speed the process way up, so you don't actually have to list 50 things. Note also that of course you could still solve this with combinatorics, but you have to be super careful as you proceed. The huge mistake people make on a question like this is calculating 5*4*5 = 100 possibilities. If you do that, you've said that the order of the boys matters, and you've therefore double counted every group. What you'd really want to do is (5C2)(5C1) = (5!/(2!3!))(5!/(4!1!)) = (5*4/2)(5) = 50.
The probability of a grouping of 2 boys and 1 girl from this group is 50/120, or 5/12.
Thanks! You the bomb! :D
ReplyDeleteThe "math" solution to #18 (Phil's cards) isn't hard if you teach students to use the multiplication rule but always handle constrained positions first. Namely: Draw four blanks for the four spots the cards could occupy. There are two choices for the first spot (because the card that goes there could be either of the two hearts), but only one remaining heart and so one choice for the last spot. After you've used those cards and spots, there are 2 remaining choices (the club or diamond) for the second spot and only 1 card/choice left for the third spot. So there are 2*1*2*1 = 4 favorable outcomes, out of 4!=24 possible permutations, and 4/24=1/6.
ReplyDeleteI don't disagree. I like listing because it's simple and not-too-slow, and the rare counting question on the SAT will have small enough numbers to make it manageable. When I work with people on the GRE, where the numbers are bigger, I use more general counting principles.
ReplyDeleteI remember being taught counting and combinatorics this way. First, we worked with problems small enough in scale to work them out the "long" way. Then, from those results, we derived the general multiplication rules. For the purposes of the SAT, I'm trying to be as inclusive as possible by not introducing any unnecessary notation, since some students might not have seen factorials yet when they're preparing for this exam, and personal experience has shown me that, whether or not the "math" solution seems simple to those of us who are well-steeped in this stuff, it often intimidates students who are very capable of getting these questions right by listing. So the decision I make is pedagogical and pragmatic, but I respect it if you instruct your students in a different way. :)
Fair point, but in my experience students often know they COULD just list the possibilities, but hate doing it (especially when there are more than ten or twelve), because they know they're missing something -- especially when I beat them over the head to recognize that if they're doing ANYTHING tedious and brute-force on the SAT, except for case-by-case analysis for number-theoretic (find the number of all ordered pairs xy such that xy=12 and x+y is odd) and digit analysis problems (if the units digit of n^6 is 9, what is the sum of the possible values of the units digit of n, they're doing it wrong and have missed the very fast, simple insight they want to test. Anyway, I commented just in case an advanced student drops by and wanted to see the more powerful solution, not to critique your teaching. You're the only other tutor I've ever seen to have really understood the higher-order themes the College Board repeatedly tests (parabolic symmetry; solving for complex algebraic expressions like sums, e.g. x+y, without solving for both x and y, which is often not even tedious but impossible; etc.)
ReplyDeleteThanks for the kind words. I think we're more similar than different, and I always love meeting new tutors who know their stuff.
ReplyDeleteIf I have to pick one problem that I think informs my tutoring decisions regarding complexity in counting, it's the plumber problem in the Blue Book. There's a question to which the 'listing' solution is very straightforward, and so is the 'math' solution. BUT there's also an incorrect math solution that seems very reasonable at first glance, and of course the result of that approach is an incorrect answer choice. I've seen so many students (and one notable teacher) miss that question using a math approach that I decided at some point to advocate the slightly less elegant but foolproof listing approach to that kind of question.
I think we should combine the 2 ways you guy mention. Under pressure, sometimes I find changing the daunting probability problem involved huge numbers (which Mike says is not often tested because SAT is calc-optional) into a simplified one so helpful. That means I use listing method first for the simple one, deduce the pattern, and be confident to apply the formal-school method for the real intimidating one. I try to balance them, and if got stuck with one, I say to myself let turn to the other. What matters here is the appropriate method one thinks himself fit perfectly and at ease to perform in real test. Also, thinking 2 ways to solve probability problems ensures that we make a right answer.
ReplyDeleteRight! It's good to be nimble and switch back and forth between methods depending on your question.
ReplyDeleteI keep getting the same answer too - using counting principle AND listing!
ReplyDeletePWN the SAT, please help! D:
Oh, and thanks!
Oh, I got it, I must have counted the same sums (such as 2+3 and 3+2) twice!
ReplyDeleteMuch thanks.
Glad you figured it out. Let me know if you run into any other issues!
ReplyDeleteI hate these kinds of problems :c it's my weakness.
ReplyDeleteSorry again to keep bombarding you with questions. :)
ReplyDeleteThere are different ways to arrange the letters E, F, and G in a row from left to right. How many more ways are there to arrange the letters D, E, F, and G in a row from left to right?
(A) 4
(B) 5
(C) 6
(D) 12
(E) 18
For this problem, I got it correct (using inductive reasoning and luck) and I want to know the "mathy" way to solve it and the "not mathy" way. The non-mathy way would probably be listing, but I figured that for the question, it could be better phrased as: "How many ways are there to arrange the letters D,E, F, and G in a row from left to right?" That question and the test question are equivalent, right? Then I just did 4x3x2, which is 24. Also, isn't it permutation since order matters? If that is the case, then isn't it 4 P 4. I don't get it.
Edit: There are SIX different ways...
ReplyDeleteWell, you've already figured out both ways, so all I need to do is confirm for you. :)
ReplyDeleteYes, this is a case of 4P4 if you prefer that notation. Or if you just use the general "how many choices do I have for this spot" counting principle, you do 4 × 3 × 2 × 1 = 24. Personally, I think being able to list out the 24 possibilities of a 4P4 quickly is a useful skill for tougher questions, but you certainly don't need to do it here.
Yeah, but isn't 4 P 4= n!/(n-r)!=4!/4-4)!
ReplyDeleteOh sorry. IGnore my naive and "stupid" comment. I forgot that 0!=1. Sorry :(
ReplyDeletehow do you solve 17??
ReplyDeleteFor each bite, there's a 1/2 probability that the snake that did it will be green. If bite 1, bite 2, bite 3, and bite 4 are all from green snakes, then it's (1/2)(1/2)(1/2)(1/2) = 1/16. Basically, I was just trying to create a problem that's slightly more interesting than a coin flip problem. This question could have just been Corey flipping a coin and getting heads 4 times in a row.
ReplyDelete