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. If A and B are two events defined on a sample space, then: \[P(A \text{ AND } B) = P(B)P(A|B) \label{eq1}\] This rule may also be written as: \[P(A|B) = \dfrac{P(A \text{ AND } B)}{P(B)} \nonumber\] (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 \[P(A|B) = P(A). \nonumber\] and Equation \ref{eq1} becomes \[P(A \text{ AND } B) = P(A)P(B). \nonumber\] If A and B are defined on a sample space, then: \[P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) \label{eq5}\] If A and B are mutually exclusive, then \[P(A \text{ AND } B) = 0. \nonumber\] and Equation \ref{eq5} becomes \[P(A \text{ OR } B) = P(A) + P(B). \nonumber\] Example \(\PageIndex{1}\) Klaus is trying to choose where to go on vacation. His two choices are: \(\text{A} = \text{New Zealand}\) and
\(\text{B} = \text{Alaska}\). 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. \(\text{A} =\) the event Carlos is successful on his first attempt. \(P(\text{A}) = 0.65\). \(\text{B} =\) the event Carlos is successful on his second attempt. \(P(\text{B}) = 0.65\). Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made
the first goal is 0.90. Solutions a. The problem is asking you to find \(P(\text{A AND B}) = P(\text{B AND A})\). Since \(P(\text{B|A}) = 0.90: P(\text{B AND A}) = P(\text{B|A})
P(\text{A}) = (0.90)(0.65) = 0.585\) Carlos makes the first and second goals with probability 0.585. b. The problem is asking you to find \(P(\text{A OR B})\). \[P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B}) = 0.65 + 0.65 - 0.585 = 0.715\] Carlos makes either the first goal or the second goal with probability 0.715. c. No, they are not, because \(P(\text{B AND A}) = 0.585\). \[P(\text{B})P(\text{A}) = (0.65)(0.65) = 0.423\] \[0.423 \neq 0.585 = P(\text{B AND A})\] So, \(P(\text{B AND A})\) is not equal to \(P(\text{B})P(\text{A})\). d. No, they are not because \(P(\text{A and B}) = 0.585\). To be mutually exclusive, \(P(\text{A AND B})\) must equal zero. Exercise \(\PageIndex{1}\) Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. \(\text{C} =\) the event that Helen makes the first shot. \(P(\text{C}) = 0.75\). \(\text{D} =\) the event Helen makes the second shot. \(P(\text{D}) = 0.75\). 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? Answer \[P(\text{D|C}) = 0.85\] \[P(\text{C AND D}) = P(\text{D AND C})\] \[P(\text{D AND C}) = P(\text{D|C})P(\text{C}) = (0.85)(0.75) = 0.6375\] Helen makes the first and second free throws with probability 0.6375. Example \(\PageIndex{2}\) 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.
Answer
Exercise \(\PageIndex{2}\) A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? Answer \[P = \dfrac{200-140-40}{200} = \dfrac{20}{200} = 0.1\] Example \(\PageIndex{3}\) Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25. Let: \(\text{M} =\) math class, \(\text{S} =\) speech class, \(\text{M|S} =\) math given speech
Answer a. 0.1625, b. 0.6875, c. No, d. No Exercise \(\PageIndex{3}\) A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student check out a DVD. Suppose that \(P(\text{B}) = 0.40, P(\text{D}) = 0.30\) and \(P(\text{D|B}) = 0.5\).
Answer
Example \(\PageIndex{4}\) Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let \(\text{B} =\) woman develops breast cancer and let \(\text{N} =\) tests negative. Suppose one woman is selected at random.
Answers
Exercise \(\PageIndex{4}\) A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports? Answer Let \(\text{A} =\) student is a senior going to college. Let \(\text{B} =\) student plays sports. \(P(\text{B}) = \dfrac{140}{200}\) \(P(\text{B|A}) = \dfrac{50}{140}\) \(P(\text{A AND B}) = P(\text{B|A})P(\text{A})\) \(P(\text{A AND B}) = (\dfrac{140}{200}\))(\(\dfrac{50}{140}) = \dfrac{1}{4}\) Example \(\PageIndex{5}\) Refer to the information in Example \(\PageIndex{4}\). \(\text{P} =\) tests positive.
Answer a. 0.98; b. 0.1401; c. 0.857; d. 0.15 Exercise \(\PageIndex{5}\) A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student checks out a DVD. Suppose that \(P(\text{B}) = 0.40, P(\text{D}) = 0.30\) and \(P(\text{D|B}) = 0.5\).
Answer
References
ReviewThe multiplication rule and the addition rule are used for computing the probability of \(\text{A}\) and \(\text{B}\), as well as the probability of \(\text{A}\) or \(\text{B}\) for two given events \(\text{A}\), \(\text{B}\) defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events \(\text{A}\) and \(\text{B}\) are mutually exclusive events when they do not have any outcomes in common. Formula ReviewThe multiplication rule: \(P(\text{A AND B}) = P(\text{A|B})P(\text{B})\) The addition rule: \(P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B})\) Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino. In this problem, let:
Suppose that one Californian is randomly selected. Exercise \(\PageIndex{5}\) Find \(P(\text{C})\). Exercise \(\PageIndex{6}\) Find \(P(\text{L})\). Answer 0.376 Exercise \(\PageIndex{7}\) Find \(P(\text{C|L})\). Exercise \(\PageIndex{8}\) In words, what is \(\text{C|L}\)? Answer \(\text{C|L}\) means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder. Exercise \(\PageIndex{9}\) Find \(P(\text{L AND C})\) Exercise \(\PageIndex{10}\) In words, what is \(\text{L AND C}\)? Answer \(\text{L AND C}\) is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder. Exercise \(\PageIndex{11}\) Are \(\text{L}\) and \(\text{C}\) independent events? Show why or why not. Exercise \(\PageIndex{12}\) Find \(P(\text{L OR C})\). Answer 0.6492 Exercise \(\PageIndex{13}\) In words, what is \(\text{L OR C}\)? Exercise \(\PageIndex{14}\) Are \(\text{L}\) and \(\text{C}\) mutually exclusive events? Show why or why not. Answer No, because \(P(\text{L AND C})\) does not equal 0. GlossaryIndependent EventsThe occurrence of one event has no effect on the probability of the occurrence of another event. Events \(\text{A}\) and \(\text{B}\) are independent if one of the following is true:
What are the laws of addition and multiplication of probability?The probability of events A and B occurring can be found by taking the probability of event A occurring and multiplying it by the probability of event B happening given that event A already happened. If events A and B are independent, simply multiply ( ) by ( ).
What is multiplication and addition theorem explain with suitable examples?If A and B are independent events associated with a random experiment, then P(A∩B) = P(A). P(B) i.e., the probability of simultaneous occurrence of two independent events is equal to the product of their probabilities. By multiplication theorem, we have P(A∩B) = P(A). P(B/A).
Is and multiplication or addition in probability?If you want the probability of A and B you multiply if the events are independent. If they are dependent you have to mutiply the probability of one by the conditional probability of the other, given the one. If you want the probability of A or B you add if the events are mutually exclusive.
How do you add and multiply probabilities?The best way to learn when to add and when to multiply is to work out as many probability problems as you can. But, in general: If you have “or” in the wording, add the probabilities. If you have “and” in the wording, multiply the probabilities.
|