## Posts tagged with: probability

Hi Mike, Test 7, Section 4, Number 38. Probability: I keep losing track trying to think this through. Other than the table provided, is there a chart I can draw that will help me “see” the probabilities? Or, what’s the best / quickest way to manage this one? Thanks !

Yeah, this chart and the paragraph explaining it are pretty confusing.

A rewording of the question would make it a lot easier to follow. The question should say: “If a contestant who received a score of 5 on any day is selected at random, what is the probability that the contestant did so on Day 2 or Day 3?”

So here’s the proportion you need to fill in:

The question tells you that no contestant received the same score on two different days. That’s important because it means if you got a score of 5 on one day, you didn’t get a score of 5 on any other day. So nobody Ken Jennings’d it—there were 7 scores of 5 in total, and 7 different people got them. So the denominator of the probability is 7—there are only 7 contestants who could be chosen at random who received a score of 5, period.

The numerator is the sum of the number that received a score of 5 on Day 2 (that’s 2 people) and the number that received a score of 5 on Day 3 (that’s 3 people). So the numerator is .

Putting it all together:

Would you please show how you would solve PSAT test section 4, #16? Thank!s

Sure! First, figure out how likely it is that a 14 or 15 year old would not have a summer job. 69 out of 89 kids ages 14-15 do not have a summer job. is approximately 0.78.

Now figure out the likelihood of a 16 or 17 year old not having a summer job. 42 out of 81 kids ages 16-17 do not have a summer job. is approximately 0.52.

How many times more likely is a 14-15 year old than a 16-17 year old NOT to have a summer job? . That’s answer choice C.

(If you don’t round in the intermediate steps, the answer still rounds neatly to 1.50: , which rounds to 1.50.)

If two dice are rolled, what is the probability the sum of the numbers is 4 or 5?
* i actually thought the two dice are the same and there is no difference between (1,4) (4,1). Then my answer was 4/36=1/9
But the answer is 7/36, apparently it considered the (1,4) (4,1) (2,3) (3,2) etc. Why should we think of the dice as different?

You have to consider those possibilities separately because, while there are two ways to get a 3 and a 1 to get a sum of 4, there’s only one way to get two 2s.

I think it’s easiest to understand this after seeing all the 36 possible rolls laid out:

Count all the 4s and 5s—there are 7 of them. If you ever get a dice probability question on the SAT (which you probably won’t), just take 20 seconds to draw that chart out, then count the cells you care about.

The figure above represents four offices that will be
assigned randomly to four employees, one employee
per office. If Karen and Tina are two of the four
employees, what is the probability that each will be
assigned an office indicated with an X ?

The figure is Four boxes or (offices) and two of them are labeled X.

See a discussion of this question from the Tumblr Q&A here.

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## How’s everyone else doing on this quiz?

15% got 5 right
31% got 4 right
23% got 3 right
12% got 2 right
18% got 0 or 1 right

Disclaimers: 1) Probability problems are some of the SAT’s most difficult, but they’re also some of the most rare. There’s a pretty decent chance you won’t see a very hard question like this on your test, so prioritize your prep time; don’t worry too much about this stuff until you’ve really nailed the basics. Ironically, this is my most involved post to date, but it corresponds to the smallest point potential. 2) As you surely know if you’ve done probability in school, as involved as this is it really only scratches the surface of a concept that can go VERY far down the rabbit’s hole. I’m only covering the kind of stuff you might see on the SAT. If you’re looking for a more complete treatment of probability, try this. 3) I’m assuming some rudimentary knowledge of combinations here.

##### 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.

1. What’s the probability of rolling an even number on a standard 6-sided die?
2. What’s the probability of rolling a number less than 6 on a standard 6-sided die?
3. What’s the probability of rolling a prime number on a standard 6-sided die?
4. What’s the probability of flipping a coin and getting either heads or tails?
5. What’s the probability of flipping two coins and having one or the other (but not both) come up heads?

OMG this is so easy! Well buckle up, kids. Things are about to get more interesting.

Yo dawg I heard you like dice.

What if, after the hot dog eating contest, there was a vomit cleaning contest, and someone from your class was going to be chosen at random to participate in that one, too. What is the probability that you’ll end up in the hot dog eating contest AND that Ashley will end up in the vomit cleaning contest? When looking for the probability of multiple events, you MULTIPLY their probabilities. You have a 1/30 chance of being chosen for the hot dog eating contest, and Ashley has a 1/30 chance of being chosen for the vomit cleaning contest. The probability that you will be chosen to eat AND she will be chosen to clean is (1/30)(1/30) = 1/900. Wow, that’s small! 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):

1. What is the probability of flipping 4 coins and having them all come up heads?
2. 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?
3. 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?

To summarize what we’ve got so far:

Probability of X or Y = (Probability of X)+(Probability of Y)*
*(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:

1. What is the probability of flipping 3 coins and having 2 of them come up heads and 1 come up tails?

(A)

(B)

(C)

(D)

(E)

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.

Possibilities (satisfactory outcomes bolded)
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

The simpler you keep a question like this, the less likely you are to make a mistake. Let’s look at one more (slightly tougher) question:

1. 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)

(B)

(C)

(D)

(E)

Straight up: it’s not easy to come up with the “math” solution. Bragging rights to whoever does so first in the comments, but I’m just going to solve it by listing possibilities. Yes, this really is the way I solve these questions when I score 2400s. 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

Take a minute to note the order in which I made my lists. If you practice listing things in order from smallest to greatest, you can get very fast at it, which makes a question like this a piece of cake. Start by “anchoring” the first 2 digits, and listing all the possible combinations of the last 2. Then anchor another set of 2 digits at the beginning, and repeat. In other words, list all the outcomes that start with “12” then all the ones that start with “13,” then all the ones that begin with “14.” Only move on to outcomes that begin with “2” once you’ve exhausted all the ones that begin with “1.” 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

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Also try this recent Weekend Challenge question involving probability (post contains multiple explanations).

Prize this week: you get to decide, once and for all, whether Marshmallow Peeps or Cadbury Creme Eggs are the better candy. No longer will disagreements be chalked up to “difference of opinion.” Once you issue your decree, it will be a matter of record. Minor dissent will be tantamount to outright prevarication. Think you can handle all that power?

Right, the question. Here we go.

Four adjacent offices are to be assigned to four employees at random. What is the probability that Scooter and The Big Man (two of the employees) will be placed next to each other?

UPDATE: Props to Elias, who nailed it on Facebook. He’s going with whichever candy has the least packaging (a man after my own sustainable heart). I’m thinking that’s the Creme Egg.

Solution below the cut…

The biggest mistake kids make with probability questions is to overcomplicate them. If you’ve done a unit on probability in school, you know things can become very complicated very quickly, and that it’s pretty easy to make a mistake under pressure. On the SAT, of course, that’s doubly true. However, on the SAT you’ll always be dealing with small numbers (remember: it’s a calculator optional test), so you can sidestep all the necessary calculations if you just get fast at LISTING COMBINATIONS!

Let’s just number the employees 1-4, and list all the possible ways they could be distributed to the four offices:

1234   2134   3124   4123
1243   2143   3142   4132
1324   2314   3214   4213
1342   2341   3241   4231
1423   2413   3412   4312
1432   2431   3421   4321
Take a minute to understand the order in which I listed those, because that’s important. I listed them from least to greatest, taking care to list all the ones that started with “12,” then all the ones that started with “13,” etc. I broke them into columns every time I had a new digit in the first position. I can check my work to make sure I’ve listed them all by making sure all the columns are the same length.
Now that I’ve listed every single possibility, I only need to count the ones that fit the question. How many have the same two people adjacent to each other? Let’s say Scooter is 1 and The Big Man is 2:
1234 2134 3124 4123
1243 2143 3142   4132
1324   2314 3214 4213
1342   2341   3241   4231
1423   2413 3412 4312
1432   2431 3421 4321
Count ’em up! Scooter and The Big Man are next to each other 12 of the 24 times! That’s a probability of 12/24, which simplifies to 1/2, or 0.5.
Remember that the numbers will stay small on the SAT. If you get fast at listing possibilities, this is probably the biggest matrix of this kind you’ll ever have to make. If you have to make one with 5 possibilities (which I’ve seen once), it’s still doable. It’s a much faster and more sure-fire way to get a question like this correct than trying to work through all the crazy probability calculations.
Ok, but what about the math? What makes this question so difficult to solve from a math perspective (and it is difficult — hundreds of you viewed this post for an average time of 3:23 without answering) is that you have a combination of ORs and ANDs:
• Scooter could get the first slot AND The Big Man could get the second slot,
• OR Scooter could get the second slot AND The Big Man could get the first slot OR the third slot,
• OR Scooter could get the third slot AND The Big Man could get the second slot OR the fourth slot,
• OR Scooter could get the fourth slot AND The Big Man could get the third slot.
Whoa. Break that down into a big ol’ probability expression (more on how to do this in my massive probability post):
So there you go. Of course, the math works out nicely as well, once you know how to set it up. I stress again: for this kind of question it is easier and less time consuming to do it the “long” way, unless you are an absolute prodigy at probabilities. When I take the test, I who PWNs the SAT and gets 2400s, I list all the possibilities and count. But hey man, it’s your life.