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.

Tried to pin the tail on the donkey

 

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.

  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) \dfrac{1}{3}
     
    (B) \dfrac{3}{8}
     
    (C) \dfrac{1}{2}
     
    (D) \dfrac{5}{8}
     
    (E) \dfrac{7}{9}

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) \dfrac{1}{2}
     
    (B) \dfrac{1}{3}
     
    (C) \dfrac{1}{4}
     
    (D) \dfrac{1}{6}
     
    (E) \dfrac{1}{3}

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).