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Lesson 138: More About Probability (Part 2 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, and the previous Lesson 137. Make sure that you fully understand these lessons, since this lesson takes the topic to the next level.
Until
now, we've been looking at probabilities involving "or" situations.
For example, the probability of rolling a 5 or a 6. In this
lesson, we'll look at the probability of "and" situations. For
example, let's say I'm going to roll a red die, and then a blue die.
I'd like to know the probability of rolling a 5 on the red, and a 3
on the blue. One way to solve this problem is by writing out
what is called the sample space of the experiment,
shown at left.
In this case, the first number in each pair is the outcome of the red die roll. The second number is the outcome of the blue die roll. You would usually label the rows and columns accordingly. From this chart, we can see that there are 36 different outcomes to this experiment. Of these, only 1 represents the outcome of a 5 red roll, and a 3 blue roll. Therefore the probability is 1/36.
How did we get 36 different outcomes? What we actually did was multiply the number of outcomes that could occur for one die (which is 6), times the number of outcomes for the second die (which is also 6). Since we're in an "and" situation, we multiply the probabilities. Recall that for "or" situations, we add.
What we have learned is called the counting principle. Using variables, we can say that if one activity can occur in x ways, and another activity can occur in y ways, the number of ways that the two activities together can occur is xy ways. We multiply the ways.
What if you were asked for the probability that the two die rolls would add up to 4? Look in the chart to see how many pairs add to 4. In this case, (1, 3) and (3, 1) would be considered different outcomes, since we're dealing with two specific dice. You can see that there are 3 different ways that we can get a sum of 4, so the probability is 3/36.
Sometimes
instead of writing out a sample space, we might be asked to draw a
tree diagram of the possible outcomes. Let's
use an example with fewer combinations. Let's say I'm going to
flip a coin, and then roll a die. What are the possible
outcomes of the two events taken together?
According to the counting principle, there will be 12 possible outcomes: two possible coin flip outcomes, times 6 possible die rolls. You can this has been diagrammed on the left. If we were asked to find the probability of a flip of heads, followed by a roll of 4, we can trace the path and see that there is 1 way out of 12, for a probability of 1/12.
A common question involving the counting principle is as follows: A man has 3 different shirts, 4 different pairs of pants, and 5 different hats. How many different outfits can he create using a shirt, a pair of pants, and a hat? This problem is very easy, as long as you don't complicate the matter. Some students ask if any of the shirts have a color clash with any of the pairs of pants, and therefore cannot go together. In these problems we assume that anything goes with anything. All we do is use the counting principle, and multiply the values given. The answer is 3 x 4 x 5 = 60 different possible outfits.
So far, in our probability problems involving "and" conditions, we've been working with independent events. That means that each event had nothing to do with the other. If you roll a die and flip a coin, each one doesn't know anything about what happened with the other. Very often, though, we deal with dependent events. That means that the outcome of the first event can have some bearing on the outcome of the second event.
A common example involves a deck of cards. If we draw two cards in a row, without replacement, what is the probability that both cards will be a diamond? We know that P(diamond) = 13/52. But now that a diamond is gone from the deck, how many diamonds are left for the second draw? There are 12 diamonds left. That means that P(diamond on second draw) = 12/51. Notice that the denominator is now 51. The first card has been drawn, and has not been put back, since we said "without replacement." To get our probability, we must multiply the two probabilities, since this is a problem involving an "and" condition. We multiply 13/52 x 12/51 and get 156/2652, which can be reduced to 1/17.
Very Important: Assuming that we are dealing with "fair" experiments (e.g., no "trick" coins), which is always how these problems are, the probability of an event occurring has nothing at all to do with what has happened in the past. For example, if a fair coin lands on heads five times in a row, the chances that it will land on heads on the sixth flip is still 1/2. The coin is not "due" to come up on tails, nor can we say that heads are "on a roll," and are more likely to come up again.
This wrong view is sometimes known as the gambler's fallacy. Each coin flip has a 1/2 chance of landing on heads. Nothing in the past matters. The coin does not have any type of "memory," nor any desire to "balance things out." With that said, certainly it is the case that over the very long haul, a coin is likely to end up on heads close to 50% of the time. But still, on any single given flip, the chances of heads are exactly 1/2. Sometimes test questions are designed to see if students understand this concept.
Remember that you can ask a math question if you have additional questions about a topic, or you can contact me if you have any comments or suggestions for this site.