Intersection of Probabilities

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I am trying to understand Probability intersections. The formula for determining the probability of events P(A) and P(B) is P(A) x P(B). However, I’m not seeing that played out in many examples. Perhaps I’m completely missing something. For example, in the screen link above, the lesson gives this exercise:

  • An advertisement company runs a quick test and shows two ads on the same web page (ad “A” and ad “B”) to 100 users. At the end of the trial, they found:
    • 12 users clicked on ad “A”
    • 17 users clicked on ad “B”
    • 3 users clicked on both ad “A” and ad “B”

The intersection of events A (clicking on ad A) and event B (clicking on ad B) is 3/100. However, why wouldn’t 12/100 multiplied by 17/100 give us the intersection?

The same goes for the example of a fair 6-sided die. What is the intersection of the following events:

  1. Rolling an odd number: {1,3,5}
  2. Rolling a number greater than 3: {4,5,6}

The answer is of course 5, which is a probability of 1/6. However, why wouldn’t you multiply the probabilities of the two events to find the intersection?

I’m hoping I haven’t made this confusing. Any help would be greatly appreciated.

P(A) * P(B) = P (A intersection B) if and only if A and B are independent events.

In the advertisement company example, there is no remark on whether clicking ad A makes the user more/less inclined to click ad B and vice versa. IE. there has been no information whether clicking one of the ads influences the user when he sees the next ad.

When it is not explicitly stated that the events are independent, check if P(A) * P(B) equals P(A intersection B) if they are not equal, then the events are dependent.

For a more intuitive example, you can think of two events X, Y
Let X be the event of you being hungry at 12 pm
Let Y be the event of you clicking a pizza ad at 12 pm
When are X, Y dependent. When are they not?

If you hate pizzas / if you never click on an ad / have an ad blocker in place, then the event of you being hungry will most likely NOT affect the probability of whether you click on the pizza ad. In this case, the events would be independent.

Otherwise, we’d assume that they are dependent, since you would be more likely to click on a pizza ad when you are hungry than when you are not.

In the case of the die example, it would be helpful to think of conditional probabilities.
P(A/B) = P(getting an odd number given that the number is greater than 3) =
= P(A intersection B)/P(B)
= 1/6 divided by 1/2
= 1/3
P(A) = P(getting an odd number) = 3/6 = 1/2

P(A/B) != P(A)
therefore, the events are not independent.

I believe this will help: https://brilliant.org/wiki/probability-independent-events/

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You can also use a Venn diagram for problems that require intersections. There are two categories, expressed as circles in the left and right. Anything that belongs to both circles goes in their intersection. If it belongs to only one, it goes in the left or right. If it belongs to neither, it goes outside. Sometimes these are represented by the quantity in each circle, the intersection, and outside instead. Here’s an example of a Venn diagram for the dice problem. There are six items, 2 is outside, 4 and 6 belong only to the right circle which is greater than 3, 1 and 3 belong only to the left circle which is odd, and 5 belongs to both circles.

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