Example: Getting two questions right on an exam

Compound, independent events

Compound Probability of Independent Events
Getting At Least One Heads
Example: Probability of rolling doubles
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Frequency Probability and Unfair Coins
Example: Getting two questions right on an exam
Example: Rolling even three times
Independent probability
Frequency Stability

Example: Getting two questions right on an exam Probability of two independent events occuring.

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Example: Getting two questions right on an exam

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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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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...

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At 2:33, Sal says "double bonds" but should say "single bonds."

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