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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.

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• 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? • Could there be a way to do this in our heads? Or without a chart? • 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!
• 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.
• How can you solve this type of question without setting up a table. What is the equation? • I don't get how to do this subject AT ALL. HEEELLLPPP! (Those who manage give me a reasonable answer gets a home-made invisible cookie and a million katroodle dollars. :D) • All the vacations seem equally fun to you, so you just decide to pick one of the three at random. That is 3 possible trips to go on.

Instead of thinking of day length of vacations, because I think that is confusing, instead think of it like this:
You can choose one friend to take with you on the vacation, either Amy, Ben, or Chris. Because you are fair, you decide to pick which friend gets to go with you at random. This means there are 9 possible ways to go on vacation:

Island with Amy, Ben, or Chris (3 possibilities): IA, IB, IC
Skiing with Amy, Ben, or Chris (3 possibilities): SA, SB, SC
Camping with Amy, Ben, or Chris (3 possibilities): CA, CB, CC

So our SAMPLE SPACE consists of these 9 total outcomes.

Here's where it gets trickier: we can set other conditions. Say that Ben does not like to ski. This means that one possible vacation, Skiing with Ben (SB), has been eliminated. How many possible vacations are left? 8, since we removed one possibility from our total SAMPLE SPACE of 9. Therefore, the probability of picking a "good" vacation (because Ben would ruin our vacation by complaining a lot) would be 8/9.

Or another example: You don't want to go camping because it's too hot. If you pick a vacation at random, what is the probability you will enjoy your vacation? Well, there are three possible outcomes that result in us going camping, so if we want to have fun, we would want to pick outcomes that are NOT camping. How many of those are there? Well, there are 9 total outcomes, minus 3 outcomes for camping, giving us 6 outcomes out of a total of 9, or 6/9.

Let's make it a little more complicated and put these two conditions together. Ben is a wimp and doesn't want to go skiing, but it's too hot to go camping. How many outcomes will satisfy us for a vacation? Now in this case, it's easier to count the situations where we would not like the vacation, so 1/9 (Skiing with Ben) and 3/9 (Camping with Anybody), so 4/9 outcomes will not satisfy us. That leaves 5 out of 9 that will.

It gets harder when the conditions overlap. Say we do not want to go skiing, OR we do not want to go with Ben. There are 3 conditions were we go with Ben, and 3 conditions where we go skiing. So the probability should be 6/9, right?

Well, no. The reason is that we are counting some outcomes more than once. We have to take out all the outcomes where they overlap. In this case, we are counting Skiing with Ben twice. So we have to take out that one condition and only count it once, leaving us with 5/9 chances.

Hope that helps! Message me if you need any more help!