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### Course: 7th grade (Illustrative Mathematics)>Unit 8

Lesson 6: Lesson 9: Multi-step experiments

# Probability of a compound event

Learn how to use sample space diagrams to find probabilities.

## Want to join the conversation?

• How can we solve this without a chart? • Khan Acamedy is pretty cool for people who are studying in this program • A jar holds 15 red pencils and 10 blue. What is the probability of picking a red pencil? • Example question;
If you roll two fair six-sided dice, what is the probability that the sum is 4 or higher?

Without drawing out a grid, what is the mathmatical formula for such a question? • The hard answer is that there really isn't one catch-all plug-and-play formula for what you want. What you're asking for is really the combination of several probability events. Let's break it down:

Let P(A) be the probability of the first die roll, and P(B) be the second die roll. A and B are discrete random variables with outcomes S = {1:6} Our sample space contains 36 possible outcomes, but not all of those outcomes are equally likely. For example, there are many ways to roll 7, but only one combination will roll 12.
So mathematically, you want this: P(A + B) >= 4.

This is a trickier question to answer than you might think, since from a mathematical perspective, here is what you are really asking:
P(A + B) = P(4 OR 5 OR 6 OR 7 OR 8 OR 9 OR 10 OR 11 OR 12)

So when you bring inequalities into the mix, you are actually asking about every single possible outcome above the one you want! Each of these probability events must be individually calculated and summed. Probability gets very complex very quickly when you start asking about probabilities beyond single events.

An easier way would be to use the complement:

P(A+B) = 1 - P(2 OR 3)

This is much easier to find. There is only one combination that gives us 2, so P(2) = 1/36. There are two possiblities for 3, 1 and 2, and 2 and 1. So P(3) = 2/36. Since these events are independent (the dice do not influence each other), we can sum the probabilities. Therefore:

P( (A+B) >= 4) = 1 - 3/36 or about 92%.

Hope that helps!
• Could there be a way to do this in our heads? Or without a chart? • there are 15 servers in a restaurant , each owns 5 identically colored shirts. each shirt chosen independently by each server each day. what is the probability that they all show up on the same day wearing the same colored shirt? • Is there a video on the probability of compound events? • What would be the mathematical way, with numbers that is, to find these probabilities, especially for the numerator, without having to draw the sample space?
For the practice problem after “Probabilities of compound events” I don’t want to have to create these tables for the coins and dices. • I do not think any one formula will cover these examples.
So you have to consider each case independently.
WIth dice, there is 1 way to get 2 or 12, 2 ways to get 3 or 11, 3 ways to get 4 or 10, 4 ways to get 5 or 9, 5 ways to get 6 or 8, and 6 ways to get 7. Note 6*6=36 and 2(1+2+3+4+5)+6=36.
For 3 coin flips, if order does not matter, having all 3 the same would be 1 (either heads or tails) and having 2 of 1 and 1 of other would be 3 (either two heads and a tail or two tails and a head) which gives 2(1+3)=8 which is 2^3=8.
If order matters, each one is separate from the others.
• I seriously need a vacation to one of those places. Upvote this post if any of you have younger siblings that are tiring. • The hard answer is that there really isn't one catch-all plug-and-play formula for what you want. What you're asking for is really the combination of several probability events. Let's break it down:

Let P(A) be the probability of the first die roll, and P(B) be the second die roll. A and B are discrete random variables with outcomes S = {1:6} Our sample space contains 36 possible outcomes, but not all of those outcomes are equally likely. For example, there are many ways to roll 7, but only one combination will roll 12.
So mathematically, you want this: P(A + B) >= 4.

This is a trickier question to answer than you might think, since from a mathematical perspective, here is what you are really asking:
P(A + B) = P(4 OR 5 OR 6 OR 7 OR 8 OR 9 OR 10 OR 11 OR 12)

So when you bring inequalities into the mix, you are actually asking about every single possible outcome above the one you want! Each of these probability events must be individually calculated and summed. Probability gets very complex very quickly when you start asking about probabilities beyond single events.

An easier way would be to use the complement:

P(A+B) = 1 - P(2 OR 3)

This is much easier to find. There is only one combination that gives us 2, so P(2) = 1/36. There are two possiblities for 3, 1 and 2, and 2 and 1. So P(3) = 2/36. Since these events are independent (the dice do not influence each other), we can sum the probabilities. Therefore:

P( (A+B) >= 4) = 1 - 3/36 or about 92%.

Hope that helps! 