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Lesson 137:More About Probability (Part 1 of 2)
In this very important lesson we'll learn more about probability. It is a topic that frequently appears on exams, and has many real life applications.
Before starting this lesson, make sure that you review Lesson 77 on probability. Make sure that you fully understand it, since this lesson takes the topic to the next level.
First we'll talk about some notation and terminology. When we write P(E), we're asking for the likelihood that an event E will occur. E will be something in particular, within the context of the problem. For example, in the context of flipping a coin once, we might ask for P(heads), which would be 1/2. In the context of a die roll, we might ask for P(2), which would be 1/6. There is a 1 in 6 chance of rolling a 2.
What are actually computing is the possible successful outcomes divided by the total possible outcomes. For example, What is P(odd) in the context of a die roll? It is 3/6, or 1/2. There are 3 possible odd outcomes, out of 6. We then reduced to lowest terms.
In probability problems, you will often see the term "fair" as in "fair die" or "fair coin." The assumption in these problems is that we're not dealing with biased or "trick" coins or dies. The chances of any outcome are equal.
Note that many probability problems involve questions about a standard deck of 52 playing cards. These questions might not be totally fair to students of a culture that does not use the Western standard deck of cards, but it's important to be fully familiar with the cards in the deck so that you can answer the questions. If anyone needs help with this, contact me and I can provide details.
A probability is always a fraction between 0 and 1. We say that the probability of an impossible event is 0, and that the probability of a certain event is 1. In the context of a deck of cards, an example of an impossible event is P(11 of clubs). There is no such card. An example of a certain event is P(red or black). Any card you draw is guaranteed to be either red or black.
Very often we deal with complementary events, or events that are complements of each other. That means events such that if one does not occur, the other definitely will. For example, in the context of a die roll, P(1 or 2) is the complement of P(3 or 4 or 5 or 6). If one of those events doesn't occur, the other will. The probabilities of complementary events always add up to 1, which is certainty. For example, P(1 or 2) = 2/6. P(3 or 4 or 5 or 6) = 4/6. They add up to 1.
If you are told that the probability that something will occur is x, then the probably that it will not occur is 1 - x, since we need for the two probabilities to add up to 1. For example, the weatherman says that the probability that it will snow tomorrow is 2/5. That means that the probability that it will not snow is 1 - 2/5, or 3/5. Review working with fractions if this is confusing.
We've already seen that when we deal with probability questions involving an "or" condition, we add the probabilities. For example, for a die roll, P(1 or 2) is 1/6 + 1/6 = 2/6. The probability of 1 is 1/6, and the probability of 2 is also 1/6. It is important to understand that this is only the case if the two possible outcomes do not overlap. For example, there is no way that the result can be both a 1 and a 2 at the same time. There is no overlap. Our formula is P(A or B) {with no overlap} = P(A) + P(B)
Compare that to this question. When drawing a card from a standard deck, what is P(7 or diamond). Here you have to be careful. We have overlap. There are four sevens in the deck, but one of them is a diamond. That means that if we add the probability of drawing a 7, plus the probability of drawing a diamond, we will have counted the seven of diamonds twice, which we can't do.
What we do is add our probabilities, but the subtract the probability that both events occurred. This compensates for the fact that we counted the 7 of diamonds twice, as part of each probability. Our formula is P(A or B) {with overlap} = P(A) + P(B) - P(A and B). Let's try it. P(7) = 4/52. There are 7 sevens. P(diamond) = 13/52. There are 13 diamonds. There is one seven of diamonds, which got counted twice. We'll subtract the probability of drawing it, which is 1/52. Our answer becomes (4 + 13 - 1) / 52, or 16/52.
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