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### Course: Precalculus (Eureka Math/EngageNY) > Unit 5

Lesson 1: Topic A: Lesson 1: The general multiplication rule- Compound probability of independent events
- Independent events example: test taking
- "At least one" probability with coin flipping
- Independent probability
- Probabilities of compound events
- Dependent probability introduction
- Dependent probability
- The general multiplication rule

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# The general multiplication rule

When we calculate probabilities involving one event AND another event occurring, we multiply their probabilities.

In some cases, the first event happening impacts the probability of the second event. We call these

**dependent events**.In other cases, the first event happening does not impact the probability of the seconds. We call these

**independent events**.## Independent events: Flipping a coin twice

What is the probability of flipping a fair coin and getting "heads" twice in a row? That is, what is the probability of getting heads on the first flip AND heads on the second flip?

Imagine we had $100$ people simulate this and flip a coin twice. On average, $50$ people would get heads on the first flip, and then $25$ of them would get heads again. So $25$ out of the original $100$ people — or $1/4$ of them — would get heads twice in a row.

The number of people we start with doesn't really matter. Theoretically, $1/2$ of the original group will get heads, and $1/2$ of that group will get heads again. To find a fraction of a fraction, we multiply.

We can represent this concept with a tree diagram like the one shown below.

We multiply the probabilities along the branches to find the overall probability of one event AND the next even occurring.

For example, the probability of getting two "tails" in a row would be:

**When two events are independent, we can say that**

*Be careful! This formula only applies to independent events.*

## Practice problem 1: Rolling dice

Suppose that we are going to roll two fair $6$ -sided dice.

## Dependent events: Drawing cards

We can use a similar strategy even when we are dealing with dependent events.

Consider drawing two cards, without replacement, from a standard deck of $52$ cards. That means we are drawing the first card, leaving it out, and then drawing the second card.

What is the probability that both cards selected are black?

Half of the $52$ cards are black, so the probability that the first card is black is $26/52$ . But the probability of getting a black card changes on the next draw, since the number of black cards and the total number of cards have both been decreased by $1$ .

Here's what the probabilities would look like in a tree diagram:

So the probability that both cards are black is:

## Practice problem 2: Picking students

A table of $5$ students has $3$ seniors and $2$ juniors. The teacher is going to pick $2$ students at random from this group to present homework solutions.

## The general multiplication rule

**For any two events, we can say that**

The vertical bar in $P(\text{B}|\text{A})$ means "given," so this could also be read as "the probability that B occurs

*given*that A has occurred."This formula says that we can multiply the probabilities of two events, but we need to take the first event into account when considering the probability of the second event.

If the events are independent, one happening doesn't impact the probability of the other, and in that case, $P(\text{B}|\text{A})=P(\text{B})$ .

## Want to join the conversation?

- How does the general multiplication rule apply to three events?(44 votes)
- Good question!

For three events A, B, and C, the extension of the general multiplication rule is

P(A and B and C) = P(A)P(B given A)P(C given (A and B)).(94 votes)

- How would you expand P(EFG|H) using the multiplication rule?(5 votes)
- It depends on what you mean by "EFG".

It could be:

- P(HHH|unfair coin)

- P(3,4,5|fair die) [3 or 4 or 5]

- ...(2 votes)

- Can someone explain how this, "A dice is thrown 5 times. If getting an even number is a success, find the probability of all 5 successes."

Equals = 1/32 ?(2 votes)- Yes.

There are 6 possible and equally likely outcomes for each die, of which 3 are even numbers.

So, the probability of rolling an even number on a die is 3∕6 = 1∕2.

Since the five dice are independent events, we can multiply their probabilities together, so the probability that all five dice show even numbers is (1∕2)⁵ = 1∕32.(8 votes)

- does term "given" , means we need to subtract the P(A) from (B). I almost understood the topic just the last formula P(B/A), makes me little uncomfortable.(4 votes)
- Not quite. P(B|A) Is the same as most of the other problems. Here is an example: You have a bag with 5 marbles. 3 of them are yellow, (Y), and 2 are green, (G). What is the probability that you will pull 2 yellow marbles in a row. This question has the pre-reqs: you are equally likely to pull any marble out of the bag, and once you pull one you do not put it back in. This means it is a
**dependent**event. the statement p(y) means the probability you will pull a yellow marble (3/5), so given, is saying What is the probability you will pull a yellow marble assuming you pulled one the first time. since they are dependent the likely hood you pull a yellow marble the 2nd time becomes 2/4, and this is the statement. P(Y|Y) so you multiply them, and the answer is P(y|y)=6/20.(1 vote)

- How we find the probability of one event out of three events when they are dependent events.(3 votes)
- When dealing with dependent events, such as drawing cards without replacement or selecting students from a table, we can still find the probability of one event out of multiple dependent events by using the general multiplication rule. This rule states that we multiply the probabilities of each event occurring given the previous events have occurred. For example, in the case of drawing two cards without replacement, if we want to find the probability of both cards being black, we first find the probability of drawing a black card on the first draw. Then, since the second draw depends on the outcome of the first draw (as the deck has changed), we find the probability of drawing a black card on the second draw given that the first card drawn was black. Multiplying these probabilities gives us the overall probability of both cards being black.(2 votes)

- I've been given an example that says: "A person has a probability of getting pulled over for speeding as 0.8; a person that is pulled over has a probability of getting a ticket as 0.9". This is a P(E and F) = P(E)*P(F|E), which comes out to .72.

What I am asking, is if the person is pulled over period, why wouldn't they have a .9 chance of getting a ticket.

I have answered my own question while typing this out. I believe that there is a .72 chance of a person getting a ticket while being pulled over for speeding due to the probablity of even getting pulled over for speeding?(3 votes) - A coin is tossed repeatedly. the coin is unfair and p(H)= p. the game ends, the first time that two consecutive heads (HH) or two consecutive tails (TT) are observed. the player wins if HH is observed and losses if TT is observed. find the probability that player wins.(2 votes)
- if a coin is flipped thrice what is the probability of obtaining atmost 2 tails . please explain by using the above theorems if possible(2 votes)
- "at most" problems are similar to "at least" problems. Instead of calculating all of the possible desired outcomes, we find the complement (undesired) and 'flip' the results to what we actually desire. Let me first explain why finding the desired is too time consuming - "at most 2 tails" means we would have to find the probability for 0 tails, 1 tail, or 2 tails and add all of them together. Instead, finding NOT "at most" only has one possibility, 3 tails. Since we know all of the possibilities together have to be 1, we can take 1 - P(3 tails)=P(at most 2 tails).

With that in mind, here's the actual process: P(3 tails)= P(1 tail)*P(1 tail)*P(1 tail) or (1/2)^3 = 1/8. Since that is the complement of what we desire, we take 1-1/8 to find our actual answer of 7/8. (You can also use a tree diagram for this problem to illustrate this, however it's a good opportunity to use the concept of complements to complete more complicated problems)(2 votes)

- Man I almost forgot that all the probability equations can be manipulated to be able to find what each other is.(2 votes)
- DO we need to learn the tree diagrams?(2 votes)