Download INDEPENDENT AND DEPENDENT PROBABILITY 1. A recent study

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INDEPENDENT AND DEPENDENT PROBABILITY
1. A recent study found that 102 of the 814 women in the survey had developed cancer in the
past ten years. What type of probability is used?
Finding probabilities of simple events, especially when the outcomes are equally likely, is quite
straightforward. But we want to determine the probability of several things happening together.
One technique is to carefully list all the ways that the experiment can turn out (the complete
sample space) and then see which outcomes favor the event.
We can determine the probability of the event by adding the probabilities of all the outcomes
favoring that event. The tricky part of this technique is to find a method of listing all possible
outcomes without forgetting or repeating any.
2. Suppose we toss two coins.
(a) What are all the possible outcomes?
(b) What is the sum of these probabilities?
(c) When can you expect to get a sum of 1 when adding probabilities?
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3. Suppose a brown paper bag carries one red ball and two green balls, identical except for
color. In a first experiment you take out one ball, note its color, and then replace it. You then
take out a second ball.
(a) Represent the sample space for the two draws.
(b) How many events are in the sample space?
4. For a second experiment, draw from the bag twice again, but this time, do NOT replace the
first ball.
(a) Once again, represent the sample space for the two draws. List the elements in the sample
space.
(b) How do the two sample spaces differ?
In the first experiment above when two balls were drawn, the probabilities for the second draw
were in no way affected by the first draw, since all the balls were available at each draw. This is
an example of independent events.
But in the second experiment when two balls were drawn, the probabilities for the second draw
were affected by the first draw, since the first ball drawn was no longer available. This is an
example of dependent events. Thus it is important to keep in mind whether such an experiment
is repeated with replacement or without replacement.
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5. A bag contains the following twelve buttons of different colors. Find the probability for each
event if the button is replaced.
(a) Pink button twice
(b) Yellow button then a green button
(c) Blue button twice
(d) Green button followed by a blue button
6. Find the probability for each event if the button is not replaced.
(a) Pink button twice
(b) Yellow button then a green button
(c) Blue button twice
(d) Green button followed by a blue button
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7. Suppose you have two circular spinners. The experiment is to spin both spinners once and
note the color you get on each. We are interested in the probability of the outcome, green on
Spinner 1 and green on Spinner 2.
Suppose that you do the experiment 1800 times. How many times do you expect to get green on
the first spinner?
Of those ________ spins having green on the first spinner, how many do you expect to give
green on the second spinner also?
(a) Using the two spinners, what is the probability of getting at least one green?
(b) Of getting no greens?
(c) Of getting at least one red?
(d) Of getting no reds?
(e) Of getting at least one blue?
(f) Of getting no blues?
(g) Of getting at most one green?
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8. Karen is signing up for a math class. She will be randomly assigned to one of six sections.
There are six sections offered:
– Three are on Tuesday and Thursday and
– Three are on Monday, Wednesday, and Friday.
– One of the Tuesday-Thursday classes and one of the Monday,
Wednesday, Fridays classes are evening sections.
What is the probability that she will be in a Tuesday- Thursday class or an evening section?
Review of Ideas
•
We say two events are independent if knowing the outcome of one event does not
change the probability of the other event occurring.
•
When two events are independent, the probability of both happening is the product of the
two probabilities.
P(A and B) = P(A) x P(B)
•
P(A and B) = 0 when the events are dependent since both events cannot occur
simultaneously.
•
P(A or B) = P(A) + P(B) – P(A and B)