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### Course: Precalculus>Unit 8

Lesson 1: Venn diagrams and the addition rule

Venn diagrams and the addition rule for probability. Created by Sal Khan.

## Want to join the conversation?

• Ok, so I have a very basic question, the answer to which seems counter intuitive to me. Ok, so if you flip a coin 4 times only, why do we multiply each possible outcome (2) by each other instead of adding them? So for instance, each flip comes with 2 possible outcomes, heads or tails. If you flip 4 times, then there should be 4X2 or 8 outcomes in my mind. But I know that to be false. There are 16, or 2*2*2*2. It seems as though it should be 2+2+2+2. I seem to conceptualize addition and multiplication does not seem correct to me. Why do we multiply? Ok thanks for any answers.
• Try thinking about the sequence of flips as follows (bear with me, and it should become clear when we get to the third flip!):

After you flip the coin once, you have 2 outcomes:

`H` (you flipped heads)
`T` (you flipped tails)

When you flip the coin a second time, you get another 2 outcomes, which (as you say) seem like they get 'added' to the previous outcomes. So now you have 4 outcomes:

`1 2` (flip number)
`H H` (first flip heads, this flip heads)
`H T` (first flip heads, this flip tails)
`T H` (first flip tails, this flip heads)
`T T` (first flip tails, this flip tails)

So far, it doesn't look like it matters whether you add or multiply, since both 2+2 and 2*2 = 4.

But now consider what happens when you flip the coin a third time. You have to 'add' another 2 outcomes to each of the previous four outcomes. So you are adding 2, four times. This is what multiplication is - multiple addition! So now there are 8 possible outcomes:

`1 2 3` (flip number)
`H H H`
`H T H`
`T H H`
`T T H`
`H H T`
`H T T`
`T H T`
`T T T`

Hopefully you can now see that if you flip a fourth time, you would need to 'add' the two new outcomes to each of the previous 8 possibilities. Adding 2 eight times is the same as 8 x 2, so there are then 16 possible outcomes. I hope this helps! :-)
• Hello everyone! I hope this question is not too hard to answer. I understand why we remove the intersection (5/29), to avoid overestimating the probability. We use 5/29, because this is a given value (we already know that there are 5 yellow cubes). But when we apply the intersection rule [P(yellow)*P(cubes)], we get: 12/29*13/29, which is not equivalent to 5/29. Why is that so? What am I missing here? Thank you!
• I take it that by the "intersection rule" you mean the rule which states:

P( A ∩ B ) = P(A) x P(B)

This rule only applies when the two events are independent. This is not always a given. What independence means is that the probability of event B is the same whether or not even A occurred.

In this case, there is (overall) a 12/29 = 0.41 chance of drawing something Yellow. However, if we know that we picked a Cube, the probability that we have something Yellow is no longer 0.41, it's 5/13 = 0.38. Hence, the probability is not constant. So the events are not independent, and we can't just multiply the probabilities to get the intersection.
• why would the probability be zero in case of mutual exclusiveness when you can count the probability that someothing or something else has been taken out of the bag even if they dont overlap? Like if you have green, red and yellow cubes and you ask about the probability of taking out green or red than you can solve that even though they dont overlap. this got me confused.
• this is just a bit of common sense. if two events cannot happen at the same time, how likely is it that they will happen at the same time? the definition of the events gives the answer to the question. it is completely unlikely that they will happen at the same time because they cannot happen at the same time!
• What if you have three or more groups that may or may not overlap, and you want to calculate P(A or B or C ... n)?
• That's a great question. As you might guess, things get very complicated pretty quickly based on how many variables there are and how well the various overlaps behave. Google "Inclusion-Exclusion Principle" to see how deep that rabbit hole goes!
• kk so i need help. Let's say i have 27 blueberry pancakes. How many banana pancakes would i need to add to make the probability of grabbing a banana pancake 10%?
(1 vote)
• Let 𝑏 be the number of banana pancakes.

Thereby we have a total of 𝑏 + 27 pancakes.

We want the probability of picking a banana pancake to be 10%:
𝑏∕(𝑏 + 27) = 0.1

Multiplying both sides by 𝑏 + 27, we get
𝑏 = 0.1𝑏 + 2.7

Subtracting 0.1𝑏 from both sides, we get
0.9𝑏 = 2.7

Finally, dividing both sides by 0.9, we get
𝑏 = 2.7∕0.9 = 3

So, we need to add 3 banana pancakes.
• Is the venn diagram necessary?
• Not required, but it explains how to "see" that you are double counting some data, and makes you "see" that you need to subtract that value once to account of the double counting.
• anyone noticed the error @ (12+13)/29-5 !=20/29!
• If I know what you mean, I believe it was actually (12 + 13 - 5)/29 = 20/29. :)
• 90% of students believed the true headline.
82% of students believed the false headline.
75% of students believed both headlines.

90+82-75 = 97% of students believed the false or the true headline.

Does 90+82 = 172/2 = 86% tell us anything? Is this the percent of students who believed one headline but not the other?

I guess what I'm really asking is, does adding up the non-intersecting parts of a pair of connected Venn diagrams and dividing by two tell us anything useful?
• Interesting question!

Suppose a student is selected at random. Define the random variable X as the fraction of the headlines that the selected student believes (so X = 0 if the student believes neither headline, X = 1/2 if the student believes exactly one headline, and X = 1 if the student believes both headlines). Then (90% + 82%)/2 = 86% is the expected value (theoretical mean) of X. In words, the expected value of the fraction of the headlines that the selected student believes is 0.86 .

Have a blessed, wonderful Christmas!