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## Precalculus

### Course: Precalculus > Unit 8

Lesson 2: Multiplication rule for probabilities- Compound probability of independent events
- Independent events example: test taking
- General multiplication rule example: independent events
- Dependent probability introduction
- General multiplication rule example: dependent events
- Probability with general multiplication rule
- Interpreting general multiplication rule
- Interpret probabilities of compound events

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# Independent events example: test taking

Have you ever taken a test and discovered you have no choice but to guess on a couple of problems? In this example problem, we are considering the probability of two independent events occurring. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- what is the difference between mutually exclusive event and independent event?(14 votes)
- Mutually exclusive:

The to events cannot happen at the same time. If you roll a die, you cannot get both an even and an odd.

Independent:

Knowledge that one event has occurred does not change the probability of another event occurring. For example, A=event that you go to work today, B=event that it rains today. If A occurs, you need to go to work, this has no impact on B, the probability of rain for that day.(56 votes)

- The easiest way I discern it is if the question includes the words "AND" vs. "OR".

"AND" means you multiply the outcomes, "OR" means you add the outcomes

( and subtract any possible overlap).

Is this too simplistic? Just curious....(18 votes)- Yes, that works pretty well. But use common sense to see if it makes sense.(5 votes)

- Sorry to bother but how can we, with absolute certainty, know that a given event is mutually exclusive or independent? Does Sal have a video on comparing both?

Anywho, awaiting your reply! Thanks!(6 votes)- 2 events are mutually exclusive when they can't happen at the same time. For example, winning and losing are mutually exclusive because you can't win and lose at the same time. Hope I answered your question!(5 votes)

- In the video Sal mentions independent events, is there such a thing as dependent events? If so, could you give a real world example. Thanks in advance.(2 votes)
- I haven't watched the video you're talking about but I'm assuming it's probability, since we're talking about events.

Yes, for example;

If I had a huge pile of sand, and I scoop a shovel out every 10 minutes. Let's say the probability of finding a rock (making this up), is 50%. A fairly rocky sand pile but this is a made up example.

Every time I get a rock, or don't get a rock, the probability of getting one the NEXT time I shovel will be higher or lower depending on how many rocks I have left. So the second event probability, the second scoop of sand, depends on the first.

Hope I helped!(7 votes)

- My question isn't necessarily about this video but about the nature of dependent / independent events.

While doing the exercises at the end of this session the following question concerning flipping a coin came up: "What is the chance of heads in the first flip, heads in the second flip and heads or tails in the third flip?".

So the question was

P(HH(H OR T)) =?

This equivalent to

P( H AND H AND ( H OR T)) = P(H) * P(H) * P(H OR T) = 1/2 * 1/2 * 1 = 1/4.

So far all good.

Although I realized that P(H OR T) = 1, I tried to reason it through.

So the first question you have to pose is: "Are H and T here dependent or independent events?". I identified them as DEPENDENT, it is either H and consequently not T or the inverse, but they obviously dependent on one another. This means that I need to use the following formula:

P(H or T) = P(H) + P(T) - P(H AND T).

P(H) = 1/2

P(T) = 1/2

but what about P(H AND T)? It seemed obvious to me that it was 0, it is either H or T but never both, which gave

P(H or T) = 1/2 + 1/2 - 0 = 1 what was in line with my previous intuition.

However from watching Sall’s video’s I had developed the idea that if two events X and Y were dependent the probability of P(X AND Y) must be bigger than zero, and if they are independent P(X AND Y) = 0. But here I seemed to have a case that although X and Y are dependent the probability of X and Y = P(X AND Y) = 0.

Were did I go astray?(3 votes)- You went astray here:

>*However from watching Sall’s video’s I had developed the idea that if two events X and Y were dependent the probability of P(X AND Y) must be bigger than zero, and if they are independent P(X AND Y) = 0*

All that dependence means is that:`P(A) ≠ P(A|B)`

. Using the definition of conditional probability here, we have:`P(A) ≠ P(A∩B) / P(B)`

. As long as this inequality is satisfied, the two events are dependent. Note that this inequality assumes*nothing*about P(A∩B) except that it is a legitimate probability, i.e.:`0 ≤ P(A∩B) ≤ 1`

.

So, if event A has non-zero probability, but P(A∩B)=0, then he two events*must*be dependent. Furthermore, since we have P(A∩B)=0 (and hence`P(A∪B) = P(A) + P(B)`

), these events are called*mutually exclusive*.(4 votes)

- How do you calculate the probability of guessing the answers of a "multiple select"/"select all that apply" problem? For example, if there were 4 choices, you could choose any combination of the 4.(3 votes)
- Have you ever heard about the "Power set"' (https://en.wikipedia.org/wiki/Power_set)? If you haven't, then here that's what that is: it's the set of all of the subsets of a certain set: let's say we have the set {1,2}: its power set would be { { }, {1}, {2}, {1,2} }. As you can see, it also includes itself and the void set. Let's now go back to the original problem: consider your "multiple select"/"select all that apply" problem as a set (I will now consider the one with 4 choices) composed by 4 elements, which represent each one of the answers: we'll therefore have set S = {A, B, C, D}. Now, let's consider a random subset G where you put all the answers you think are correct, for example {C, D}. Your question can be now converted into "How many subsets G could I take?": well, that's the amount of elements contained in the power set. The particularity of the power set is that its number of elements is equal to 2^n, where n is the number of elements of the set you're taking the power set of. Why 2^n? Well, you could say that for each one of the elements, you have 2 choices: take it into your subset G or reject it: applying this to each element will have 2^n combinations. So your answer is 2^4, which is 16.

In this case, though, you might want to consider the void set as invalid (because that would mean that no one of the answers would be correct) and your answer will be 16 - 1 - 15.

Hope this helps, if you didn't understand something of my explanation (I suck at explaining things) just reply.(3 votes)

- Is this where matrices begin to intersect probability then, within the context of this course? Or was that more the way he just happened to illustrate this particular problem?(2 votes)
- You are right; he just did it as an illustration. I don’t think matrices are used any where else in this unit.(4 votes)

- How can I find probability of guessing answer on problem 1 OR on problem 2?

I should add the individual probabilities? If yes, why does it work?(2 votes)- I will assume that "OR" means one or the other or both (i.e. at least one). This is what "OR" usually means in math.

Adding the individual probabilities does not quite work, because this double-counts the probability of guessing both questions correctly. We must also subtract the probability of guessing both correctly.

From the lesson, the probability of guessing both correctly is 1/4 * 1/3 = 1/12.

So the probability of guessing question 1 OR question 2 correctly (or both) is 1/4 + 1/3 - 1/12 = 1/2.

An alternative method is to find the probability of guessing both incorrectly, then subtract from 1.

The probability is 3/4 of answering question 1 incorrectly, and 2/3 of answering question 2 incorrectly. Because the guesses are independent, the probability of guessing both incorrectly is 3/4 * 2/3 = 1/2.

So the probability of guessing question 1 correctly OR question 2 correctly (or both) is 1 - 1/2 = 1/2.(4 votes)

- In a similar question to the problem, what would be the probability of guessing the first question wrong but then the second question right?

Also, what would you do to find out the probability of guessing the first question wrong, then the second question wrong, then the third question correctly?(2 votes)- Let's say that they are 4-choice questions:

For 2 questions, there are 16 possible outcomes.

The 1st wrong, 2nd right can be done in 3*1 = 3 ways.

The probability of that happening is 3/16

Your second questions can be solved in the same manner(3 votes)

- What happens if the first flip doesn't affect the second flip?(2 votes)
- The video discusses test-taking and the probability of selecting two correct answers on a test randomly. The video does not discuss coin flips, but if you are interested in a coin flip example:

If the first flip does not, in any way, impact the probabilities for results of the second flip, then the two events are considered to be independent. The probability of two independent events both occurring (e.g., flipping Heads**and**Heads, or Tails**and**Tails) can be calculated by multiplying the probabilities of each independent event.

Therefore, if you wanted the probability of, for example, flipping Heads**and**Heads, then you would calculate:

P(H) = 1/2 = 0.5

P(H**and**H) = 1/2 * 1/2 = 0.5 * 0.5 = 1/4 = 0.25(3 votes)

## Video transcript

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.