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Lesson 3: Compound events and sample spaces

# Probability of a compound event

Learn how to use sample space diagrams to find probabilities.

## Want to join the conversation?

• I seriously need a vacation to one of those places. Upvote this post if any of you have younger siblings that are tiring.
• No upvote lol! I will not upvote anyone >:)
• 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?
• think of it as a fraction. There are 15 pencils out of 25 pencils so the probability of picking a red pencil would be 15/25 or 60%.
• 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?
• You Could but you are more likely to miss something
• Very helpful, but makes me want a vacation