Introduction to Statistics course :

here is the question

1-Two Basic Rules of Probability

note* The solution from this pictures

3.3 | Two Basic Rules of Probability When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. Chapter 3 | Probability Topics 189 The Multiplication Rule If A and B are two events defined on a sample space, then: P(A AND B) = P(B)P(A|B). This rule may also be written as: P(AIB) = P(A AND B) P(B) (The probability of A given B equals the probability of A and B divided by the probability of B.) If A and B are independent, then P(A|B) = P(A). Then P(A AND B) = P(A|B)P(B) becomes P(A AND B) = P(A)P(B). The Addition Rule If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) - P(A AND B). If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) - P(A AND B) becomes P(A OR B) = P(A) + P(B).Example 3.14 Open with Klaus is trying to choose where to go on vacation. His two choices are; A = New Zealand and B = Alaska * Klaus can only afford one vacation. The probability that he chooses A is P(A) - 0.6 and the probability that he chooses B is P(B) - 0.35. P(A AND B) = 0 because Klaus can only afford to take one vacation Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05, Example 3.15 Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90. a. What is the probability that he makes both goals? Solution 3.15 a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B A) = 0.90: P(B AND A) = P(BJA) P(A) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585. b. What is the probability that Carlos makes either the first goal or the second goal? Solution 3.15 b. The problem is asking you to find P(A OR B). P(A OR B) = P(A) + P(B) - P(A AND B) = 0.65 + 0.65 - 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715. c. Are A and B independent? Solution 3.15 C. No, they are not, because P(B AND A) = 0.585.P(B)P(A) = (0.65)(0.65) = 0.423 0.423 / 0.585 = P(B AND A) So, P(B AND A) is not equal to P(B) P(A). d. Are A and B mutually exclusive? Solution 3.15 d. No, they are not because P(A and B) = 0.585. To be mutually exclusive, P(A AND B) must equal zero.Try It Open with 3.15 Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event Helen makes the second shot. P(D) 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws? Example 3.16 A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty- seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. a. What is the probability that the member is a novice swimmer? Solution 3.16 28 150 b. What is the probability that the member practices four times a week? Solution 3.16 b. 80 150 c. What is the probability that the member is an advanced swimmer and practices four times a week? Solution 3.16 40 150 d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? Solution 3.16 d. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced