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Probability: Living With The Odds. 7. Combining Probabilities. And Probability: Independent Events. Two events are independent if the outcome of one does not affect the probability of the other event. If two independent events A and B have individual
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Probability: Living With The Odds 7 Combining Probabilities
And Probability: Independent Events Two events are independent if the outcome of one does not affect the probability of the other event. If two independent events A and B have individual probabilities P(A) and P(B), the probability that A and B occur together is P(A and B) = P(A) • P(B). This principle can be extended to any number of independent events.
Example What is the probability of rolling three 6s in a row with a single die? Solution
And Probability: Dependent Events Two events are dependent if the outcome of one affects the probability of the other event. The probability that dependent events A and B occur together is P(A and B) = P(A) • P(B given A) where P(B given A) is the probability of event B given the occurrence of event A. This principle can be extended to any number of dependent events.
Example A three-person jury must be selected at random from a pool that has 6 men and 6 women. What is the probability of selecting an all-male jury? Solution The probability that the first juror is male is 6/12. If the first juror is male, the remaining pool has 5 men among 11 people. The probability that the second juror is also male is 5/11 and so on.
Either/Or Probabilities:Non-Overlapping Events Two events are non-overlapping if they cannot occur together, like the outcome of a coin toss, as shown to the right. For non-overlapping events A and B, the probability that either A or B occurs is shown below. P(A or B) = P(A) + P(B) This principle can be extended to any number of non-overlappingevents.
Example Suppose you roll a single die. What is the probability of rolling either a 2 or a 3? Solution The probability of rolling either a 2 or a 3 is 1/3.
Either/Or Probabilities:Overlapping Events Two events are overlapping if they can occur together, like the outcome of picking a queen or a club, as shown to the right. For overlapping events A and B, the probability that either A or B occurs is shown below. P(A or B) = P(A) + P(B) – P(A and B) This principle can be extended to any number of overlapping events.
Example What is the probability that in a standard shuffled deck of cards you will draw a queen or a club? Solution These are overlapping events. P(A or B) = P(A) + P(B) – P(A and B) P(Queen or club) = P(Q) + P(club) – P(Q and club) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13
The At Least Once Rule(For Independent Events) Suppose the probability of an event A occurring in one trial is P(A). If all trials are independent, the probability that event A occurs at least once in n trials is shown below. P(at least one event A in n trials) = 1 –P(not event A in n trials) = 1 – [P(not A in one trial)]n
Example Use the at least once rule to find the probability of at least one head when you toss three coins. Solution In this case, the event A is getting heads on one toss, and for three coins n = 3. Therefore, the at least once rule tells us that P(at least one H in 3 tosses) = 1 – P(no H in 3 tosses) = 1 – [P(no H in 1 toss)]3
Example (cont) The probability of getting no heads in one toss is ½, so our final result is
Example You purchase 10 lottery tickets, for which the probability of winning some type of prize on a single ticket is 1 in 10. What is the probability that you will have at least one winning ticket among the 10 tickets? Solution The probability of winning with any one ticket is 0.1, the probability of not winning with one ticket is 1 − 0.1 = 0.9.
Example The probability of winning at least once with ten tickets is P(at least one win with 10 tickets) = 1 – [P(not winning1]10 = 1 − [0.9]10 ≈ 0.651