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## Statistics and probability

### Course: Statistics and probability>Unit 7

Lesson 8: Multiplication rule for dependent events

# Independent & dependent probability

This time around we're not going to tell you whether we're working on a dependent or independent probability event problem. You tell us! Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Does Probability and Statistic are from the same family? In their use I mean •   Short version of my answer is Yes. If we were to categorize math concepts into a limited number of labels (say 20 labels), probability and statistics would be categorized in the same family. The reason lies on the fact that both of them are dealing with "uncertainty".

In probability, you have uncertain events in future. What you do is:
1- build a model (in example of flipping a coin, your probability model includes one variable with two possible values: H or T, and the variable will take on each of these two values with the same likelihood)
2- try to "predict the FUTURE" : meaning that what events will occur with what likelihood.
So, in a sense, you want to start from your "model" and reach some "data".

In statistics, you're still dealing with uncertainty. Because you don't have access to the whole population, only to a small subset of it (sample). what you do is:
1- You have the data. Using statistical tools, you try to build the best probability model that describes your data.
2- With that model in hand, you want to reason about the whole population, or in a sense, you want to "look backward" and identify the whole population that your sample is a part of.
So, you start from your "data", and you move toward a "model", that best describes your data.
• Does anybody know what is mutually exclusive and independent event for venn diagram??? • In the case of independent events(A and B),
P(A|B) = P(A)
Could someone please explain the logic behind this? • Independent events have no effect on each other. If I have a deck of cards and a coin, the probability I draw a heart out of the deck of cards is not influenced by whether I had flipped a “heads” or “tails” prior to drawing the card. Likewise, the probability I flip a “heads” on the coin is not influenced by whether or not I drew a heart out of the deck prior to flipping the coin. The events of flipping the coin and drawing a card are independent of each other.

If event A is getting a “heads” by flipping a coin and event B is drawing a heart out of a deck of cards. The probability of getting a “heads” P(A) is no different than the probability of getting a “heads” given I have drawn a heart out of the desk first P(A|B). They are both a 0.5 probability. P(A)= P(A|B) for independent events.

The “given” event in the P(A|B) should be treated as though it has already happened – even if the probability of the “given” event is extremely rare. The probability of getting a “heads” given that you won the lottery is no different than the probability of getting a “heads” given that you did not win the lottery is the same as the probability of getting a "heads" before you know the lottery results. I hope that makes sense. This is true because the results of the lottery have no effect on the results of the coin flip – they are independent events.
• I'm very confused about interdependence. In many of the question's hints we see the logic:

"P(A | B)=P(A) and P(B | A)=P(B) therefore events A and B are independent".

...or similar arguments. But the results above depend on the distribution of the balls, or cards, or whatever is under evaluation. Surely we should say that if we pick a subset of all possible events, and then pick a second subset of events from the first subset, the answer is always dependent regardless of whether "P(A | B)=P(A)" or not?

From mathisfun "Example: taking colored marbles from a bag: as you take each marble there are less marbles left in the bag, so the probabilities change. We call those Dependent Events, because what happens depends on what happened before"

Or how about his. I challenge you to the following game. We both put \$1 on the table. We have a bag with 2 blue and 2 red, 4 balls altogether in it. I take out 2 balls look at it but don't show it to you. And I guess the color of the next ball. If I'm right then I win the \$2 otherwise I lose. Is that fair? No. If I pulled out a ball of each color, the odds are 50-50, but if I pulled out 2 balls of the same color then I'm guaranteed to win. In the long term I will win and therefore the events are no independent regardless of the distribution of the balls. Or am I missing something? • This difficulty (which is giving me headache as well) is in the contrived little stories the problem writers are trying to make up to humanize the issue and they're shooting us all in the foots. We're never going to pull marbles out of bags at a circus for money and very few of us will become professional card counters in Las Vegas or the French Riviera anyway so let's leave that behind.

Imagine we are the creators of the situation, a system, a little computer game. Or no wait, imagine we each already created our games and then we swapped, now we're examining a system a classmate created.
Here's what I have so far (just starting thinking this way so anyone ahead of me and doing well with this please chime in):
Two events or behaviors within the system can be seen to be independent if the probability of one of them happening is unaffected by changes made to the other. In shorthand code: Independent is when P(A|B)=P(A). In human words A is going to do whatever it does regardless of what B does. If one is causing or interfering with the other it's dependent and you can spy that out by noticing that P(A|B) CHANGES from P(A).
And so on.
Hope this helps.
• What is the difference between mutually exclusive and independent events? • What does independent and dependent mean? • Is that true? P(A)*P(B)=P(A&B) where events A and B are independent like flipping a coin twice to get 2 heads? ..then how can we judge this:
if 5 of 7 students forgot lunch, then, P(choose 2 student neither forgot lunch) = (2/7)(1/6)?
these 2 events are not independent and despite that we multiply their probabilities of the 2 events to get the probability of both together. • The 1/6 is not the probablility that the second student doesn’t forget his lunch. Rather, this is the conditional probability that the second student does not forget his lunch given that the first student does not forget his lunch. (The unconditional probability that the second student doesn’t forget his lunch, with no knowledge of the outcome for the first student, is actually 2/7.)

Whether events A and B are independent or not, it is always true that P(A and B)=P(A)P(B given A) as long as P(A) is nonzero. In the special case that A and B are independent, P(B given A) simplifies to P(B) in this equation.
• as in football games there are larger number of tickets than just 3. doesn't that huge number (i guess it will be in tens of thousands) allow us to consider the two events as independent even we do not put the first pulled ticket back?
i think later in this series you mentioned something about that, is this applicable for the example in subject? • Can you think of dependent and independent like positive and negative. For example in the video clip the first event was independent however, because the second event was dependent then both must be classified as dependent, or the result is dependent. (Negative) If they were both independent or positive they would still be positive or independent. However, if you have two dependent events or negative, does that make it independent or positive>
(1 vote) • That's an interesting connection to draw, but nope, in a sequence of events they are either all independent or all dependent. You might think about it like this:

Does the result of any event depend on any other events? Or in other terms: Are we likely to get a different result based on some of the other results we've observed? If so, these are dependent events. All of them. 