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Independent Events Lesson 6.2.2. Starter 6.2.2 State in writing whether each of these pairs of events are disjoint. Justify your answer. If the events.

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Presentation on theme: "Independent Events Lesson 6.2.2. Starter 6.2.2 State in writing whether each of these pairs of events are disjoint. Justify your answer. If the events."— Presentation transcript:

1 Independent Events Lesson 6.2.2

2 Starter 6.2.2 State in writing whether each of these pairs of events are disjoint. Justify your answer. If the events are disjoint, calculate P(A or B) –A single fair die is rolled. Event A is that an even number comes up. Event B is that an odd number comes up. –Two fair coins are tossed. Event A is that the first coin comes up heads. Event B is that the second coin comes up heads. –One card is drawn from a deck of 52 playing cards. Event A is that the card is a heart. Event B is that the card is a face card.

3 Objectives Describe what it means to say that two events are independent. Use the multiplication rule for independent events to answer “and” probability questions. Answer “at least” probability questions through the use of complementary probability.

4 Independent Events Two (or more) events are considered to be independent if the fact that one event occurs does not change the probability that the other will occur. If two events A and B are independent, then: P(A and B) = P(A) x P(B) –This is called the “independent and” rule For example, the probability of flipping two coins and getting two heads is: P(heads and heads) = (.5)(.5) =.25

5 Example A bag contains 3 red marbles and 7 black marbles. One marble is drawn and its color noted. It is then put back in the bag. Another marble is drawn and its color noted. –Event A is: the first marble is red –Event B is: the second marble is red Are the two events independent? Yes: –P(A) = 0.3. –Because the marble was replaced, P(B) is also 0.3. –The success or failure of event A had no effect on the probability of success of event B –So P(A and B) = P(both red) = (.3)(.3) =.09

6 Example A bag contains 3 red marbles and 7 black marbles. One marble is drawn and its color noted. It is then set aside. Another marble is drawn and its color noted. –Event A is: the first marble is red –Event B is: the second marble is red Are the two events independent? No: P(A) = 0.3 but P(B) is no longer 0.3 because the marble was not replaced. –The success or failure of event A changes the probability of success of event B –So we cannot compute P(A and B) by just multiplying. We will do “conditional and” computations later.

7 Calculating “at least” Probabilities If 3 coins (a quarter, a nickel and a dime) are tossed, what is the probability that at least one coin comes up heads? –Event A: the quarter is heads –Event B: the nickel is heads –Event C: the dime is heads It seems we are interested in P(A or B or C), so we just add: P(A) + P(B) + P(C) =.5 +.5 +.5 = 1.5 !!! What’s wrong? –The events are not disjoint, so the addition rule does not apply

8 Answer the question empirically Use the calculator’s Probability Simulator to toss a coin 3 times, hoping to get AT LEAST one head. Do 10 more trials, keeping track of successes and failures. –Report to me the number of successes out of 11 trials. –I will combine class results to get our empirical probability. Next, let’s approach the problem theoretically:

9 Using Complements to Answer “at least” Questions Stand the question on its head: –What is the probability of getting no heads? So we are asking the complementary question –But that means we got all three coins to come up tails, so what is the probability of 3 tails? –Now we can use the “and” rule because the coins are independent. P(tails and tails and tails) =.5 x.5 x.5 =.125 –Then the probability we want is the complement of the probability we just found, so P(at least one head) = 1 –.125 =.875

10 Another Example Three coins are tossed. Calculate the probability of getting exactly two heads by application of the “and” and “or” rules. –Hint: Write the sample space to see how many ways there are to win this bet. Then do the math. –P(H and H and T) = (.5)(.5)(.5) =.125 –P(H and T and H) = (.5)(.5)(.5) =.125 –P(T and H and H) = (.5)(.5)(.5) =.125 –These are disjoint events, so add:.125 +.125 +.125 =.375 –Note that this is the decimal equivalent of 3/8 and that there are 3 out of 8 branches on your diagram that have exactly two heads

11 Objectives Describe what it means to say that two events are independent Use the multiplication rule for independent events to answer “and” probability questions A “at least” probability questions through the use of complementary probability

12 Homework Read pages 331 – 335 Do problems 24 – 27


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