The term event is also often used in conjunction with sports. For example, the gymnastics event will be held at the Worthington Arena.

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1 Show Me: Probability M7253 The word event in ordinary English has a different meaning from the word event in mathematics. The term event usually means a special occurrence. For example, the day you were born was a special event. The term event is also often used in conjunction with sports. For example, the gymnastics event will be held at the Worthington Arena. In this lesson, we ll talk about events and how they are related to probability. There are six possible outcomes when rolling a number cube. Any one of six numbers may be rolled: one, two, three, four, five, and six. An event is an outcome or a group of outcomes. When all outcomes are equally likely, use this formula to find the probability of an event. The probability of an event equals the number of favorable outcomes divided by the total number of possible outcomes. Let s find the probability of rolling an even number on a number cube.

2 The possible outcomes are rolling a one, two, three, four, five, or six as shown in this tree diagram. There are six possible outcomes. The favorable outcomes are rolling a two, four, or six. There are three favorable outcomes. To find the probability of rolling an even number, divide the number of favorable outcomes by the total number of possible outcomes. Then simplify if necessary. Here s a question. What is the probability of the spinner landing on red? A combination of two or more events is called a compound event. To find the probability of a compound event, first determine whether the events are independent or dependent. Let s look at independent events first. Two events are independent if the occurrence of one event does not affect the occurrence of the other event. For example, rolling a number cube does not affect the tossing of a coin, so these events are independent. Are these two events independent? Event one, spinning yellow on the spinner shown. Event two, rolling a five on a number cube.

3 Let s find the probability of rolling a four on a number cube and getting tails on a coin toss. When there is more than one event, it is helpful to draw a tree diagram to list all the possible outcomes. We can think about adding a coin toss to each branch of the number cube tree diagram we ve used before. A tree diagram has one branch for each possible outcome. There are twelve outcomes possible when rolling a number cube and flipping a coin. We are trying to find the probability of rolling a four on a number cube and getting tails on a coin toss, so only one outcome, four T, is favorable. Now we ll use the probability formula to calculate the probability of rolling a four and tossing tails. The probability is one-twelfth. Luckily, there is a shorter way to calculate the probability of two independent events. We can use multiplication. For independent events A and B, the probability of both A and B occurring equals the probability of A times the probability of B. The probability of rolling a four is one-sixth, and the probability of flipping tails is onehalf. The probability of rolling a four and flipping tails is one-sixth times one-half, or onetwelfth.

4 Here s an example. Let s find the probability of spinning yellow or orange on the spinner shown and rolling an even number on a number cube. Event A is spinning yellow or orange on the spinner shown. Event B is rolling an even number on a number cube. Fill in the blanks. Click Solution to see how to solve this problem. There are two possibilities of out five to get yellow or orange on the spinner shown. So, the probability of A equals two-fifths. There are three even numbers on a number cube. So, the probability of B equals three-sixths or one-half. Because A and B are independent, the probability of both A and B occurring is the probability of A times the probability of B. So, the probability of A and B is one-fifth. Now it s time to try one. Event A is rolling a number less than four on a number cube. Event B is flipping heads on a coin toss.

5 What is the probability of A, the probability of B, and the probability of both A and B? Drawing marbles from a bag is another interesting scenario. Consider a bag of marbles that contains two blue marbles, three green marbles, four red marbles, and three purple marbles. In total, there are twelve marbles in the bag. Suppose you pull a marble out of the bag and replace it. Then you pull out a second marble. These two events are independent. You pull out a marble at random. What is the probability that the marble is red? Now you put the red marble back in the bag. The probability that a second marble drawn out of the bag is red is also four-twelfths or one-third. Because you replaced the first marble, pulling out a second red marble is an independent event. Therefore, the probability of drawing two red marbles with replacement is equal to the probability of drawing the first red marble times the probability of drawing the second red marble. So the probability of drawing two red marbles with replacement equals one-third times one-third, which equals one-ninth. Now let s think about what happens if you pull a marble out of the bag and do not replace it.

6 If you pull out a red marble and do not replace it, then when you go to pull out a second marble, there will only be eleven marbles left in the bag. And only three of them are red. So, the probability that you draw a red marble next, if you did not replace the first red marble, is three-elevenths. In this case, the two events are dependent events. The outcome of the second event depends on what happened in the first event. By not replacing the first marble, the total number of marbles in the bag and the types of marbles available changed from what was originally in the bag. Remember, two events are dependent events if the outcome of one affects the outcome of the other. True or false. You pull a blue marble out of the bag and you do not replace it. You pull out a second marble. The probability of drawing a second blue marble is the same as the probability that the first marble drawn was blue. Let s look at an example. There were two blue marbles, five green marbles, and seven brown marbles in a bag. You pulled out a blue marble and did not replace it. Find the probability of pulling out a green marble next. Click Solution for the correct answers. You started with fourteen marbles. You drew out one blue marble and did not replace it, so there are thirteen marbles left in the bag. One blue marble, five green marbles, and seven brown marbles. Five of the

7 thirteen marbles left are green. The probability of drawing a green marble next out of the remaining marbles is five-thirteenths. Here s one to try. There were two red marbles, three blue marbles and five gold marbles in a bag. You pulled out one red marble and did not replace it. Find the probability of drawing a gold marble next. Suppose you have a locker with a three-digit combination. Let s check out three-digit locker combinations where all three nonzero digits are divisible by three. There are nine nonzero digits. There are three nonzero digits that are divisible by three three, six, and nine. The probability that the first nonzero digit is divisible by three is three-ninths which equals one third. The probability that the second nonzero digit is divisible by three is three-ninths which equals one third. The probability that the third nonzero digit is divisible by three is also three-ninths which equals one third. What if we add the restriction that no digits can be repeated in the locker combination? Then, the digits of the lock are no longer independent. If the first digit is a three, the second and third digits cannot be a three. The probability that the first nonzero digit is divisible by three is still equal to three-ninths or one-third. Now, to consider the probability that the second digit is also divisible by three, consider the new restriction that no digits may repeat. So, there is one less nonzero digit that is

8 divisible by three and there is also one less nonzero digit available to consider for the total number of digits possible. Now, there are only two nonzero digits divisible by three out of eight possible nonzero digits. So, the probability that the second nonzero digit is divisible by three is equal to two-eights or one-fourth. By the time we reach the third digit, there is only one digit left that is divisible by three. What is the probability that the third nonzero digit is divisible by three? The key ideas of this lesson are: Two events are independent if the outcome of one event has no effect on the outcome of the second event. Two events are dependent if the outcome of one event affects the outcome of the second event. The probability of independent events can be found using a tree diagram or the formula probability of A and B equals the probability of A times the probability of B. The probability of a second dependent event happening after a first event has already happened can be found by considering the effects of the first event. If you d like to review this activity again, click Review. If you re ready to exit, click Done.