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Current time:0:00Total duration:4:44

AP.STATS:

VAR‑4 (EU)

, VAR‑4.E (LO)

, VAR‑4.E.1 (EK)

, VAR‑4.E.2 (EK)

On a multiple choice test,
problem 1 has 4 choices, and problem 2 has 3 choices. That should be choices. Each problem has only
one correct answer. What is the probability of
randomly guessing the correct answer on both problems? Now, the probability of guessing
the correct answer on each problem-- these are
independent events. So let's write this down. The probability of correct
on problem number 1 is independent. Or let me write it this way. Probability of correct on number
1 and probability of correct on number 2, on problem
2, are independent. Which means that the outcome
of one of the events, of guessing on the first problem,
isn't going to affect the probability of guessing
correctly on the second problem. Independent events. So the probability of guessing
on both of them-- so that means that the probability of
being correct-- on guessing correct on 1 and number 2 is
going to be equal to the product of these
probabilities. And we're going to see why that
is visually in a second. But it's going to be the
probability of correct on number 1 times the probability
of being correct on number 2. Now, what are each of
these probabilities? On number 1, there are 4
choices, there are 4 possible outcomes, and only one of them
is going to be correct. Each one only has one
correct answer. So the probability of being
correct on problem 1 is 1/4. And then the probability of
being correct on problem number 2-- problem number 2 has
three choices, so there's three possible outcomes. And there's only one correct
one, so only one of them are correct. So probability of correct
on number 2 is 1/3. Probability of guessing correct
on number 1 is 1/4. The probability of doing on both
of them is going to be its product. So it's going to be equal to
1/4 times 1/3 is 1/12. Now, to see kind of visually
why this make sense, let's draw a little chart here. And we did a similar thing for
when we thought about rolling two separate dice. So let's think about
problem number 1. Problem number 1 has
4 choices, only one of which is correct. So let's write-- so
it has 4 choices. So it has 1-- let's write
incorrect choice 1, incorrect choice 2, incorrect choice 3,
and then it has the correct choice over there. So those are the 4 choices. They're not going to necessarily
be in that order on the exam, but we can just
list them in this order. Now problem number 2
has 3 choices, only one of which is correct. So problem number 2 has
incorrect choice 1, incorrect choice 2, and then let's say the
third choice is correct. It's not necessarily in that
order, but we know it has 2 incorrect and 1 correct
choices. Now, what are all of the
different possible outcomes? We can draw a little
bit of a grid here. All of these possible
outcomes. Let's draw all of
the outcomes. Each of these cells or each of
these boxes in a grid are a possible outcome. You could-- you're
just guessing. You're randomly choosing one
of these 4, you're randomly choosing one of these 4. So you might get incorrect
choice 1 and incorrect choice 1-- incorrect choice in problem
number 1 and then incorrect choice in
problem number 2. That would be that
cell right there. Maybe you get this-- maybe you
get problem number 1 correct, but you get incorrect choice
number 2 in problem number 2. So these would represent all of
the possible outcomes when you guess on each problem. And which of these outcomes
represent getting correct on both? Well, getting correct on both
is only this one, correct on choice 1 and correct on choice--
on problem number 2. And so that's one of the
possible outcomes and how many total outcomes are there? There's 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, out of 12 possible outcomes. Or since these are independent
events, you can multiply. You see that they're 12 outcomes
because there's 12 possible outcomes. So there's 4 possible outcomes
for problem number 1, times the 3 possible outcomes for
problem number 2, and that's also where you get a 12.

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