Answer: (-4, 0) and (0, -2). Bell Work: Find the x and y intercepts of the following equation, then graph. x + 2y = -4.

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Presentation transcript:

Bell Work: Find the x and y intercepts of the following equation, then graph. x + 2y = -4

Answer: (-4, 0) and (0, -2)

Lesson 83: Probability of Dependent Events

We know that when events are independent, the probability that both will occur is the product of their probabilities. P(A and B) = P(A)P(B)

In this lesson, we will compute probabilities of events that are not independent. Events that are not independent are sometimes called dependent events because the probability of one event depends on the other event.

Example: Two eighth graders and one seventh grader want to go on the trip, but only two students will be selected. If their names are drawn at random, what is the probability that the two eighth graders will get to go? (make a tree)

Answer: Make a tree. 1/3 chance

Rather than counting outcomes in the first example, we can consider probabilities at each stage of the experiment. In the first stage (the selection of the first student), 2/3 of the students are eight graders.

After an eighth grader is selected, there is a 50% chance (or probability of ½) that the second eight grader will be selected. The probability that two eighth graders are selected is ⅔½ which is ⅓.

This illustrates a multiplication rule for events A and B that are not independent. P(A) under initial conditions  P(B) under new conditions

The previous problem is an example of Conditional Probability The previous problem is an example of Conditional Probability. The selection of the first eighth grader and the second are not independent events. When one name is selected, the probability of selecting the next name changes.

This differs from independent events in which one event does not change the probability of future events.

For example, consider an experiment in which a marble is selected from a bag containing 2 white and 3 blue marbles, its color is recorder, and it is replaced. Then another marble is drawn, recorded, and replaced. The events, “blue” (for the first marble) and “blue” (for the second marble), are independent because every time a marble is selected the probability of selecting a blue marble remains 3/5.

However, if a blue marble is drawn and not replace, the conditions have changed and the probability of selecting another blue marble changes to ½.

Example: A bag contains two white marbles and three blue marbles. If two marbles are selected, what is the probability of selecting two blue marbles if the marbles are selected With replacement Without replacement

Answer: 3/5  3/5 = 9/25 3/5  2/4 = 6/20 = 3/10

HW: Lesson 83 #1-25