How does the general multiplication rule apply to three events?

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)).

How would you expand P(EFG|H) using the multiplication rule?

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] - ...

if a coin is flipped thrice what is the probability of obtaining atmost 2 tails . please explain by using the above theorems if possible

"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)

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 ?

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.

Main content

Course: Statistics and probability > Unit 7

Lesson 8: Multiplication rule for dependent events

The general multiplication rule

Google Classroom

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:

P (T and T) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

When two events are independent, we can say that

P (A and B) = P (A) \cdot P (B)

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.

problem 1

Find the probability that both dice show a $3$ ‍.

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:

P (both black) = \frac{26}{52} \cdot \frac{25}{51} \approx 0.245

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.

problem 2

Find the probability that both students selected are juniors.

The general multiplication rule

For any two events, we can say that

P (A and B) = P (A) \cdot P (B | A)

The vertical bar in

P (B | 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 (B | A) = P (B)

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Steve
Posted 7 years ago. Direct link to Steve's post “How does the general mult...”
How does the general multiplication rule apply to three events?
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(43 votes)
Answer
- Ian Pulizzotto
  Posted 7 years ago. Direct link to Ian Pulizzotto's post “Good question! For three...”
  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)).
  Comment on Ian Pulizzotto's post “Good question! For three...”
  (85 votes)
Kristin
Posted 6 years ago. Direct link to Kristin's post “How would you expand P(EF...”
How would you expand P(EFG|H) using the multiplication rule?
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(5 votes)
Answer
- Juliano Penna
  Posted 3 years ago. Direct link to Juliano Penna's post “It depends on what you me...”
  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]
  - ...
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  (2 votes)
guncha.ch
Posted 5 years ago. Direct link to guncha.ch's post “does term "given" , means...”
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.
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(4 votes)
Answer
- Gabi Adaline
  Posted 8 days ago. Direct link to Gabi Adaline's post “Not quite. P(B|A) Is the ...”
  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.
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  (1 vote)
Maxim Lifirenko
Posted 5 years ago. Direct link to Maxim Lifirenko's post “Can someone explain how t...”
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 ?
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(1 vote)
Answer
- Jerry Nilsson
  Posted 5 years ago. Direct link to Jerry Nilsson's post “Yes. There are 6 possibl...”
  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.
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  (7 votes)
Rian
Posted 3 years ago. Direct link to Rian's post “I've been given an exampl...”
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?
Button navigates to signup pageComment on Rian's post “I've been given an exampl...”
(3 votes)
Answer
sweta jha
Posted 5 years ago. Direct link to sweta jha's post “A coin is tossed repeated...”
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.
Button navigates to signup pageComment on sweta jha's post “A coin is tossed repeated...”
(2 votes)
Answer
meridelsa2
Posted 5 years ago. Direct link to meridelsa2's post “if a coin is flipped thri...”
if a coin is flipped thrice what is the probability of obtaining atmost 2 tails . please explain by using the above theorems if possible
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(2 votes)
Answer
- Joe Tesoriero
  Posted 3 years ago. Direct link to Joe Tesoriero's post “"at most" problems are si...”
  "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)
  Button navigates to signup page
  (2 votes)
Suaeb Ahmed
Posted 9 months ago. Direct link to Suaeb Ahmed's post “Very good lesson!”
Very good lesson!
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(2 votes)
Answer
junejoiqra.6
Posted 2 years ago. Direct link to junejoiqra.6's post “How we find the probabili...”
How we find the probability of one event out of three events when they are dependent events.
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(2 votes)
Answer
- daniella
  Posted 2 months ago. Direct link to daniella's post “When dealing with depende...”
  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.
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  (1 vote)
Rayhaan Khan
Posted 8 months ago. Direct link to Rayhaan Khan's post “DO we need to learn the t...”
DO we need to learn the tree diagrams?
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(2 votes)
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