PROBABILITY
| Problem: | A spinner has 4 equal sectors colored yellow, blue, green and red. What are the chances of landing on blue after spinning the spinner? What are the chances of landing on red? |   |
| Solution: | The chances of landing on blue are 1 in 4, or one fourth. |
| | The chances of landing on red are 1 in 4, or one fourth. |
This problem asked us to find some probabilities involving a spinner. Let's look at some definitions and examples from the problem above.
| Definition | Example |
| An experiment is a situation involving chance or probability that leads to results called outcomes. | In the problem above, the experiment is spinning the spinner. |
| An outcome is the result of a single trial of an experiment. | The possible outcomes are landing on yellow, blue, green or red. |
| An event is one or more outcomes of an experiment. | One event of this experiment is landing on blue. |
| Probability is the measure of how likely an event is. | The probability of landing on blue is one fourth. |
In order to measure probabilities, mathematicians have devised the following formula for finding the probability of an event.
| Probability Of An Event |
| P(A) = | The Number Of Ways Event A Can Occur | | The Total Number Of Possible Outcomes | |
The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. Let's take a look at a slight modification of the problem from the top of the page.
| Experiment 1: | A spinner has 4 equal sectors colored yellow, blue, green and red. After spinning the spinner, what is the probability of landing on each color? | | |
| Outcomes: | The possible outcomes of this experiment are yellow, blue, green, and red. |
| Probabilities: | | P(yellow) | = | number of ways to land on yellow | = | 1 | | total number of colors | 4 | | | | P(blue) | = | number of ways to land on blue | = | 1 | | total number of colors | 4 | | | | P(green) | = | number of ways to land on green | = | 1 | | total number of colors | 4 | | | | P(red) | = | number of ways to land on red | = | 1 | | total number of colors | 4 | |
| Experiment 2: | A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number? |  |
| Outcomes: | The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6. |
| Probabilities: | | P(1) | = | number of ways to roll a 1 | = | 1 | | total number of sides | 6 | | | | P(2) | = | number of ways to roll a 2 | = | 1 | | total number of sides | 6 | | | | P(3) | = | number of ways to roll a 3 | = | 1 | | total number of sides | 6 | | | | P(4) | = | number of ways to roll a 4 | = | 1 | | total number of sides | 6 | | | | P(5) | = | number of ways to roll a 5 | = | 1 | | total number of sides | 6 | | | | P(6) | = | number of ways to roll a 6 | = | 1 | | total number of sides | 6 | | | | P(even) | = | # ways to roll an even number | = | 3 | = | 1 | | total number of sides | 6 | 2 | | | | P(odd) | = | # ways to roll an odd number | = | 3 | = | 1 | | total number of sides | 6 | 2 | | |
Experiment 2 illustrates the difference between an outcome and an event. A single outcome of this experiment is rolling a 1, or rolling a 2, or rolling a 3, etc. Rolling an even number (2, 4 or 6) is an event, and rolling an odd number (1, 3 or 5) is also an event.
In Experiment 1 the probability of each outcome is always the same. The probability of landing on each color of the spinner is always one fourth. In Experiment 2, the probability of rolling each number on the die is always one sixth. In both of these experiments, the outcomes are
equally likely to occur. Let's look at an experiment in which the outcomes are not equally likely.
| Experiment 3: | A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble? | | ![[IMAGE]](http://www.mathgoodies.com/lessons/vol6/images/marbles_jar.gif) |
| Outcomes: | The possible outcomes of this experiment are red, green, blue and yellow. |
| Probabilities: | | P(red) | = | number of ways to choose red | = | 6 | = | 3 | | total number of marbles | 22 | 11 | | | | P(green) | = | number of ways to choose green | = | 5 | | total number of marbles | 22 | | | | P(blue) | = | number of ways to choose blue | = | 8 | = | 4 | | total number of marbles | 22 | 11 | | | | P(yellow) | = | number of ways to choose yellow | = | 3 | | total number of marbles | 22 | |
The outcomes in this experiment are not equally likely to occur. You are more likely to choose a blue marble than any other color. You are least likely to choose a yellow marble.
| Experiment 4: | Choose a number at random from 1 to 5. What is the probability of each outcome? What is the probability that the number chosen is even? What is the probability that the number chosen is odd? |
| Outcomes: | The possible outcomes of this experiment are 1, 2, 3, 4 and 5. |
| Probabilities: | | P(1) | = | number of ways to choose a 1 | = | 1 | | total number of numbers | 5 | | | | P(2) | = | number of ways to choose a 2 | = | 1 | | total number of numbers | 5 | | | | P(3) | = | number of ways to choose a 3 | = | 1 | | total number of numbers | 5 | | | | P(4) | = | number of ways to choose a 4 | = | 1 | | total number of numbers | 5 | | | | P(5) | = | number of ways to choose a 5 | = | 1 | | total number of numbers | 5 | | | | P(even) | = | number of ways to choose an even number | = | 2 | | total number of numbers | 5 | | | | P(odd) | = | number of ways to choose an odd number | = | 3 | | total number of numbers | 5 | |
The outcomes 1, 2, 3, 4 and 5 are equally likely to occur as a result of this experiment. However, the events even and odd are not equally likely to occur, since there are 3 odd numbers and only 2 even numbers from 1 to 5.
| Summary: | The probability of an event is the measure of the chance that the event will occur as a result of an experiment. The probability of an event A is the number of ways event A can occur divided by the total number of possible outcomes. The probability of an event A, symbolized by P(A), is a number between 0 and 1, inclusive, that measures the likelihood of an event in the following way: - If P(A) > P(B) then event A is more likely to occur than event B.
- If P(A) = P(B) then events A and B are equally likely to occur.
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CONDITIONAL PROBABILITY
| Problem: | A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test? | ![[IMAGE]](http://www.mathgoodies.com/lessons/vol6/images/testpaper1.gif) |
| Analysis: | This problem describes a conditional probability since it asks us to find the probability that the second test was passed given that the first test was passed. In the last lesson, the notation for conditional probability was used in the statement of Multiplication Rule 2. |
| Multiplication Rule 2: | When two events, A and B, are dependent, the probability of both occurring is: |
| |  |
The formula for the Conditional Probability of an event can be derived from Multiplication Rule 2 as follows:
 | Start with Multiplication Rule 2. |
|
 | Divide both sides of equation by P(A). |
|
 | Cancel P(A)s on right-hand side of equation. |
|
 | Commute the equation. |
 | We have derived the formula for conditional probability. |
Now we can use this formula to solve the problem at the top of the page.
| Problem: | A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test? | |
| Solution: | | P(Second|First) | = | P(First and Second) | = | 0.25 | = | 0.60 | = | 60% | | P(First) | 0.42 | |
Let's look at some other problems in which we are asked to find a conditional probability.
| Example 1: | A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black? |
| Solution: | | P(White|Black) | = | P(Black and White) | = | 0.34 | = | 0.72 | = | 72% | | P(Black) | 0.47 | |
| Example 2: | The probability that it is Friday and that a student is absent is 0.03. Since there are 5 school days in a week, the probability that it is Friday is 0.2. What is the probability that a student is absent given that today is Friday? |
| Solution: | | P(Absent|Friday) | = | P(Friday and Absent) | = | 0.03 | = | 0.15 | = | 15% | | P(Friday) | 0.2 | |
| Example 3: | At Kennedy Middle School, the probability that a student takes Technology and Spanish is 0.087. The probability that a student takes Technology is 0.68. What is the probability that a student takes Spanish given that the student is taking Technology? |
| Solution: |
| | P(Spanish|Technology) | = | P(Technology and Spanish) | = | 0.087 | = | 0.13 | = | 13% | | P(Technology) | 0.68 | |
| Summary: | The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A), read as the probability of B given A. The formula for conditional probability is: |
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| | The Venn Diagram below illustrates P(A), P(B), and P(A and B). What two sections would have to be divided to find P(B|A)? Answer |
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| | ![[IMAGE]](http://www.mathgoodies.com/lessons/vol6/images/conditional_venn.gif) |
Moderation Team
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