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?
|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?
Possibilities (satisfactory outcomes bolded)
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:
- 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?
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).
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).
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
- 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?
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?
- 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?
Also try this recent Weekend Challenge question involving probability (post contains multiple explanations).