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## AP®︎/College Statistics

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

Lesson 2: Mutually exclusive events and unions of events

AP.STATS:
VAR‑4 (EU)
,
VAR‑4.C (LO)
,
VAR‑4.C.2 (EK)
,
VAR‑4.E (LO)
,
VAR‑4.E.4 (EK)
CCSS.Math:
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. • 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)? • 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.
• anyone noticed the error @ (12+13)/29-5 !=20/29! • Is the venn diagram necessary? • How can all the possibilities be equally likely ( ), if there are different numbers/colors of cubes/spheres? Am I misunderstanding sth? :( • Sal is talking about the fact that every single object has the same chance of falling out first. So each of the 29 objects could fall out of the bag and every single one of them is equally likely at the start of the experiment. In this case for every single object it would be 1/29. The groups(cubes, yellow,...) on the other hand have, because there are different amounts of them, a different chance of falling out. It's important to remember that even we have grouped these objects by shape and colour, they still are single, and in some sense, unique things with there very own likelihood of falling out. And the likelihood is the same for each of them.
• I am a little confused on the "or" rule for combining probabilities. Here is the problem. I have one die with six sides. If I roll the die 8 times, what is the probability that it will come up a six at least once? Using the "or" rule the formula would be
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 8/6 or a probability of 133%
Am I applying the "or" rule incorrect? Why?

How does this differ from this example. Say the odds of getting killed in a car accident are 1 death per 100,000 miles driven. Each day I drive 1,000. What is the probability that I will be die in a car accident in 50 days? (50%?) 100 days? (100%?) 200 days (200%)... • The short answer is that there is overlap on the Venn diagram and, therefore, you need to subtract that overlap to avoid double counting.
Roll 1 die: Chance of rolling a 6 is 1/6
Roll 2 dice: Chance of rolling AT LEAST ONE 6 is (chance of rolling 6 on first die) + (chance of rolling 6 on second die) - (chance of rolling 6 on BOTH dice) = 1/6 + 1/6 - 1/36 = 7/36

As the number of dice goes up, subtracting out all the possible overlaps gets really tedious (maybe you roll 6 on 3 of the dice, maybe on 5 of them, etc.)

An easier way in this case is to "flip" the question:
Instead of asking "what's the probability of rolling at least one six?" we ask "What's the probability of never rolling a six?"
P(at least one 6) = 1 - P(no 6's)
P(no 6s) = P(no 6 on first die)P(no 6 on second die)...
We can simply multiply these "AND" conditions because there is no overlap.

Thus, we have,
P(at least one 6) = 1 - P(no 6s)
= 1 - (5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)
for eight dice.

This "trick" gives the same answer as adding all the P(roll 6)'s then subtracting out all the possible overlaps, it's just a lot less error-prone and tedious.
(1 vote)
• At , why does it say on the bottom right, in a type box, 29 is also the denominator of -5? 