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### Course: AP®︎/College Statistics > Unit 7

Lesson 4: Independent versus dependent events and the multiplication rule- 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
- The general multiplication rule
- Probability with general multiplication rule
- "At least one" probability with coin flipping
- Probabilities involving "at least one" success
- Probability of "at least one" success

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# General multiplication rule example: independent events

We can use the general multiplication rule to find the probability that two events both occur when the events are independent. Created by Sal Khan.

## Want to join the conversation?

- I am curious why we cannot answer the question like the following.

The probability of getting silk on the first spin is 1/6.

The probability of getting silk on the second spin is 1/6.

So the probability that Doug or Maya will get silk is 2/6, or 1/3.

So then the probability of neither of them getting silk must be the inverse of this or 2/3, or 0.66666666666?

However the probability obtained by simply multiplying 5/6(5/6) is 25/36 or 0.6944444444444. Almost the same but not exactly.

So, it seems that we can't simply inverse the probability of the positive event to get the negative event? Why not?(11 votes)- Interesting question!

“Inversing” the “positive” event to get the “negative” event is not the mistake. Instead, the mistake is assuming that the probability of the “positive” event of either Doug or Maya getting silk is 1/6 + 1/6. This is incorrect because these two events of silk for Doug and silk for Maya could**both**occur (they are not mutually exclusive). So this calculation counts twice, instead of once, the probability that both events occur. So we must correct this by subtracting the probability that both occur, which is 1/6 * 1/6 = 1/36.

So the probability of the “positive” event is 1/6 + 1/6 - 1/36 = 11/36. Then the probability of the “negative” event is indeed 1 - 11/36 = 25/36, which matches 5/6 * 5/6.

Have a blessed, wonderful Christmas!(30 votes)

- Probability that Maya and Doug both get silk should be 1/6 x 1/6 = 1/36.

The probability that they both don't get silk is 25/36.

Shouldn't probability that they both get silk and they both don't get silk be 1 but 1/36 + 25/36 is not equal to 1. Please clarify.(1 vote)- They do not add up to 1 because there are other possibilities you have not taken into account. Maya could get silk and Doug does not, which is 1/6 x 5/6 = 5/36. Doug could get silk and Maya does not, which is also 1/6 x 5/6 = 5/36.

probability that neither gets silk + probability that only Maya gets silk + probability that only Doug gets silk + probability that both get silk

= 25/36 + 5/36 + 5/36 + 1/36

= 36/36

= 1(18 votes)

- Why is the second draw not 4/5 assuming silk was not drawn in the first draw? That is, why is one of the options not taken off the wheel after the first spin? The question isn’t very clear that both could draw the same material.(5 votes)
- The problem did not indicate that the option can be taken off the wheel after Maya's turn. If that was the case then Doug's turn would be considered as dependent to the 1st event (ie. Maya's turn) as the wheel has been altered. For the purpose of this problem, it is important that the wheel is fair and unaltered in the sequence of events.(3 votes)

- I'm curious won't Doug have total options of materials = 5, since Maya already chose one(Not silk). So won't it be 5/6 multiply by 4/5, which results in 2/3?(3 votes)
- There is no rule that they must have different materials. They could very well both get wood, or both get plastic, and so on.(3 votes)

- why the condition P(A).P(B/A), is just equal of P(B) is it because the events are dependant or independent?(3 votes)
- If P(B|A) and P(B) are equal, then the events are independent. This is because the outcome of A does not affect the probability of B.(2 votes)

- If this were dependent so that when a person got an outcome, it could not be used again, how would the expression be written?(3 votes)
- What you are referring is known as hypergeometric distribution or the generalisation which is multihypergeometric distribution.

Assuming you have understood binomial distribution you should be able to easily understand it. The probability is given by:

probability = num_events * prob_of_single_event

This works because probability of each event given they have have same composition is the same. Let R be red balls and B be blue balls. The balls selected without replacement. Then P(RRRBB) = P(RBRBR) since each event has 3 R's and 2 B's.(2 votes)

- What happens if the probability is both theoretical and experimental?(2 votes)
- They are mutually exclusive, two different things. While there is a possibility that they are equal, they are two different things. Theoretical is what should happen, experimental is what happens when you run an experiment.(2 votes)

- what does it mean by generation of multiplication(2 votes)
- If 𝐴 and 𝐵 are independent events, then 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)⋅𝑃(𝐵)

The general multiplication rule works for both independent and dependent events:

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)⋅𝑃(𝐵 | 𝐴)(2 votes)

## Video transcript

- [Instructor] We're
told that Maya and Doug are finalists in a crafting competition. For the final round,
each of them spin a wheel to determine what star material
must be in their craft. Maya and Doug both want to get
silk as their star material. Maya will spin first, followed by Doug. What is the probability that
neither contestant gets silk? Pause this video and think
through this on your own before we work through this together. All right, so first let's think
about what they're asking. They want to figure out the
probability that neither gets silk, so I'm gonna
write this in shorthand. So I'm going to use MNS for Maya no silk. And we're also thinking about Doug not being able to pick silk. So Maya no silk and Doug no silk. So we know that this could be viewed as the probability that
Maya doesn't get silk. She, after all does get
to spin this wheel first, and then we can multiply that by the probability that
Doug doesn't get silk, Doug no silk, given that
Maya did not get silk. Maya no silk. Now it's important to think
about whether Doug's probability is independent or dependent on
whether Maya got silk or not. So let's remember Maya will
spin first, but it's not like if she picks silk, that somehow silk is taken out of the running. In fact, no matter what she picks, it's not taken out of the running. Doug will then spin it again. And so these are really
two independent events, and so the probability
that Doug doesn't get silk given that Maya doesn't get silk, this is going to be the same thing as the probability that
just Doug doesn't get silk. It doesn't matter what happens to Maya. And so what are each of these? Well, this is all going to be equal to the probability that
Maya does not get silk. There's six pieces or six options of this wheel right over here. Five of them entail her not
getting silk on her spin. So five over six. And then similarly, when
Doug goes to spin this wheel there are six possibilities. Five of them are showing
that he does not get silk, Doug no silk. So times 5/6, which is of
course going to be equal to 25/36, and we're done.