Statistics: 5.3 homework
About 13% of the population of a large country is allergic to pollen. If two people are randomly selected, what is the probability both are allergic to pollen? What is the probability at least one is allergic to pollen Assume the events are independent. (a) The probability that both will be allergic to pollen is (b) The probability that at least one person is allergic to pollen is
(a) 0.0169 Both P(E)=0.13 and P(F)=0.13 P(E and F)=P(E)•P(F) P(E and F)=P(E)•P(F)=0.0169 (b) 0.2431 P(E), (PEc)=1−P(E) 1-0.13=0.87 The probability that 1 person is not computer illiterate is 0.87 P(E and F)=P(E)•P(F) P(neither E)=P(person 1 not E)•P(person 2 not E) =0.87•0.87 =0.7569 compute the probability that at least one person is computer illiterate. P(at least 1 computer illiterate)=1−P(neither computer illiterate) =1-0.7569 =0.2431
For a parallel structure of identical components, the system can succeed if at least one of the components succeeds. Assume that components fail independently of each other and that each component has a 0.14 probability of failure. (a) Would it be unusual to observe one component fail? Two components? (b) What is the probability that a parallel structure with 2 identical components will succeed? (c) How many components would be needed in the structure so that the probability the system will succeed is greater than 0.9998?
(a) It would not be unusual to observe one component fail, since the probability that one component fails, 0.09, is greater than 0.05. It would be unusual to observe two components fail, since the probability that two components fail, 0.0081, is less than 0.05. An event with a probability of less than 0.05 (or 5%) is typically considered unusual. To determine whether it would be unusual to observe one component fail, identify the probability that one component fails. The probability that one component fails is 0.14 Therefore, it would not be unusual to observe one component fail, since the probability that one component fails is greater than 0.05. To determine whether it would be unusual to observe two components fail, identify the probability that two components fail. The Multiplication Rule for Independent Events states that if E and F are independent events, then P(E and F) is given by the formula below. P(E and F)=P(E)*P(F) Recall that the probability that each component fails is 0.14 and that components fail independently of each other. Calculate the probability that two components fail. P(2 components fail)=P(first fails)*P(second fails) =0.14*0.14 =0.0196 The probability that two components fail is less than 0.05. Therefore, it would be unusual to observe two components fail. (b) 0.9804 The Complement Rule states that if E represents any event and Ec represents the complement of E, then PEc is given by the formula below. P(Ec)=1-P(E) A parallel structure with 2 identical components will succeed if at least one of the components succeeds. Since the complement of "at least one component succeeding" is "2 components fail," the probability that a parallel structure with 2 identical components will succeed can be calculated by using the Complement Rule and calculating 1−P(2 components fail). Recall from part (a) that P(2 components fail)=0.0196. Use this information to calculate the probability that at least one component succeeds. P(at least one component succeeds)=1−P(2 components fail) =1−0.0196 =0.9804 Therefore, the probability that a parallel structure with 2 identical components will succeed is 0.9804 (c) 5 The Multiplication Rule for n Independent Events states that if events E1, E2, E3, ..., En are independent, then P(E1 and E2 and E3 and ... and En) is given by the formula below. P(E1 and E2 and E3 and ... and En)=P(E1)*P(E2)*...*P(En) Recall that the system will succeed if at least one of the components succeeds. Let n represent the number of components in the system, and let Ei be the event that component i fails. An expression that represents the probability that the system will succeed is 1−[P(E1)•P(E2)•P(E3)•...•P(En)] Therefore, the number of components needed so that the probability the system will succeed is greater than 0.9998 is the smallest value of n that makes 1-[P(E1)•P(E2)•P(E3)•...•P(En)] greater than 0.9998 Recall from part (a) that each component has a 0.14 probability of failure. Therefore, 1-[P(E1)•P(E2)•P(E3)•...•P(En)]=1−(0.14 )^n. While the smallest value of n that makes 1−(0.14)^n greater than 0.9998 could be found using algebra or trial and error, for the purpose of this explanation, trial and error will be used. Recall from part (b) that the probability that a parallel structure with 2 identical components will succeed is 0.9804. This probability is not greater than 0.9998 Therefore, more than 2 components are needed. Calculate the probability that the system will succeed with 3 components. P(system succeeds)=1−(0.14)^3 =0.997256 This probability is not greater than 0.9998. Therefore, more than 3 components are needed. Continue finding the probability that a system with one additional component succeeds until the probability that the system succeeds for a given number of components is greater than 0.9998. The number of components that would be needed in the structure so that the probability the system will succeed is greater than 0.9998 is 5.
What is the probability of obtaining seven tails in a row when flipping a coin? Interpret this probability.
The probability of obtaining seven tails in a row when flipping a coin is .00781 P(tail)=1/2 P(E and F and G and...)= P(E)*P(F)*P(G)...* P(tails on all seven flips)= P(tail on 1st flip)*...*P(tail on 7th flip) =1/2*1/2*1/2*1/2*1/2*1/2*1/2 P = (1/2)^7 P = 0.00781 P(E) = relative frequency of E=frequency of E/number of trials of experiment Assume that E is the event of obtaining seven tails in a row when flipping a coin. If the number of trials of the experiment is 10,000, the expected frequency of E is 10,000•P(E). Interpret the probability P(E)=0.00781 Consider the event of a coin being flipped seven times. If that event is repeated ten thousand different times, it is expected that the event would result in seven tails about 10,000•0.00781= 78 times
The word or in probability implies that we use the _____ Rule.
addition
The word and in probability implies that we use the ________ rule.
multiplication
According to a poll, about 18% of adults in a country bet on professional sports. Data indicates that 48.2% of the adult population in this country is male. (a) Are the events "male" and "bet on professional sports" mutually exclusive? Explain. (b) Assuming that betting is independent of gender, compute the probability that an adult from this country selected at random is a male and bets on professional sports. (c) Using the result in part (b), compute the probability that an adult from this country selected at random is male or bets on professional sports. (d) The poll data indicated that 11.2% of adults in this country are males and bet on professional sports. What does this indicate about the assumption in part (b)? (e) How will the information in part (d) affect the probability you computed in part (c)? Select the correct choice below and fill in any answer boxes within your choice.
(a) No. A person can be both male and bet on professional sports at the same time. Mutually exclusive events have no outcomes in common. A person can be both male and bet on professional sports at the same time. Therefore, the events are not mutually exclusive. (b) P(male and bets on professional sports)=0.0868 P(E and F)=P(E)•P(F) P(male)=0.482 P(bets on professional sports)=0.18 P(male and bets on professional sports)=P(male)•P(bets on professional sports) =(0.482)•(0.18) =0.0868 (8.68%) (c) P(male or bets on professional sports)= 0.5752 P(E or F)=P(E)+P(F)−P(E and F) P(male or bets on professional sports)=0.482+0.18- 0.0868 =0.5752 (57.52%) (d) The assumption was incorrect and the events are not independent. What is P(male and bets on professional sports) according to the poll? P(male and bets on professional sports)=0.112 Notice that this is not equal to the value found in part (b). Because the poll determined a different value for the probability that an adult is male and bets on professional sports than was found in part (b) when the events were assumed to be independent, the assumption was incorrect and the events are not independent. (e) P(males or bets on professional sports)=0.5500 Because the events are not independent, the value calculated in part (b) cannot be used in the calculations for part (c). Find P(male or bets on professional sports). Use P(male and bets on professional sports)=0.112 instead of the value you found in part (b). P(male or bets on professional sports) =0.482+0.18−0.112 =0.5500 (55.00%)
In airline applications, failure of a component can result in catastrophe. As a result, many airline components utilize something called triple modular redundancy. This means that a critical component has two backup components that may be utilized should the initial component fail. Suppose a certain critical airline component has a probability of failure of 0.0059 and the system that utilizes the component is part of a triple modular redundancy. (a) Assuming each component's failure/success is independent of the others, what is the probability all three components fail, resulting in disaster for the flight? (b) What is the probability at least one of the components does not fail?
(a) The probability is 0.00000021 P(E1 and E2 and E3)=P(E1)•P(E2)•P(E3)=(0.0059)(0.0059)(0.0059) =0.00000021 (b) The probability is 0.99999979 P(E1 and E2 and E3)=0.00000021 1−0.00000021=0.99999979
Determine if the following statement is true or false. When two events are disjoint, they are also independent.
False *Two events are disjoint if they have no outcomes in common. In other words, the events are disjoint if, knowing that one of the events occurs, we know the other event did not occur. Independence means that one event occurring does not affect the probability of the other event occurring. Therefore, knowing two events are disjoint means that the events are not independent.
For the fiscal year 2007, a tax authority audited 1.66% of individual tax returns with income of $100,000 or more. Suppose this percentage stays the same for the current tax year. What is the probability that two randomly selected returns with income of $100,000 or more will be audited?
The probability is 0.000276 P(E and F)=P(E)•P(F) Assume that E is the event that the first randomly selected return with income of $100,000 or more will be audited and F is the event that the second randomly selected return with income of $100,000 or more will be audited. Note that these events are independent. The probability that two randomly selected returns with income of $100,000 or more will be audited is 0.0166*0.0166=0.000276 1.66/100=0.0166
The probability that a randomly selected 4-year-old male salamander will live to be 5 years old is 0.99545. (a) What is the probability that two randomly selected 4-year-old male salamanders will live to be 5 years old? (b) What is the probability that seven randomly selected 4-year-old male salamanders will live to be 5 years old? (c) What is the probability that at least one of seven randomly selected 4-year-old male salamanders will not live to be 5 years old? Would it be unusual if at least one of seven randomly selected 4-year-old male salamanders did not live to be 5 years old?
(a) The probability that two randomly selected 4-year-old male salamanders will live to be 5 years old is 0.99092 P(E)=0.99545 P(F)=0.99545 P(E and F)=P(E)•P(F)=0.99092 (b) The probability that seven randomly selected 4-year-old male salamanders will live to be 5 years old is 0.96858 How many independent events are involved in this situation? n=7 P(E1 and E2 and ... and E4)=(0.99545)^7=0.96858 (c) The probability that at least one of seven randomly selected 4-year-old male salamanders will not live to be 5 years old is 0.03142; Yes, because the probability of this happening is less than 0.05 P(Ec)=1−P(E) 1-0.96858=0.03142 Typically, an event with a probability less than 0.05 (or 5%) is considered unusual. Use this information to determine if it would be unusual that at least one of seven randomly selected 4-year-old male salamanders did not live to be 5 years old. 0.03142 is smaller than 0.05 so it would be unusual
Suppose George loses 31% of all staring contests. (a) What is the probability that George loses two staring contests in a row? (b) What is the probability that George loses four staring contests in a row? (c) When events are independent, their complements are independent as well. Use this result to determine the probability that George loses four staring contests in a row, but does not lose five in a row.
(a) The probability that George loses two staring contests in a row is 0.0961 Begin by converting the percentage of staring contests that George wins into a probability. To convert a percent into a probability, remove the percent sign and then divide by 100. George wins staring contests with a probability of 0.31 It is safe assume that the outcomes of the probability experiment are independent, because there is no indication that George's winning one staring contest affects his winning any other staring contest. Since the outcomes are independent, use the multiplication rule for independent events, which states that if E and F are independent events, then P(E and F)=P(E)•P(F) In this situation, let E be the event that George wins one staring contest and let F be the event that Nate wins another, different staring contest. In this case, P(E)=P(F) because both events are that George wins a randomly selected staring contest. In this situation, let E be the event that George wins one staring contest and let F be the event that George wins another, different staring contest. In this case, P(E)=P(F) because both events are that George wins a randomly selected staring contest. Find the probability that George wins two staring contests. P(George wins two staring contests)=0.31•0.31=0.0961 (b) The probability that George loses four staring contests in a row is 0.0092 If events E1,E2,E3,..., En are independent, then P(E1 and E2 and E3 and ... and En)=P(E1)•P(E2)•P(E3)•...•P(En) Recall that P(George wins a staring contest)=0.31 and that all of the events in this situation are the same. Substitute this value into the expression and multiply, rounding to four decimal places. 0.31*0.31*0.31*0.31=0.0092 Thus, the probability that George wins four staring contests in a row is 0.0092 (c)The probability that George loses four staring contests in a row, but does not lose five in a row is 0.0063 If E represents any event and EC represents the complement of E, then P(E^C)=1−P(E) Determine P(George does not win a staring contest) by subtracting P(George wins a staring contest) from 1 1-0.31=0.69 Use this probability, as well as the result from part (b), to determine P(George wins five staring contests in a row but does not win six in a row), rounding to four decimal places. 0.0092•0.69=0.0063 Thus, the probability that George wins five staring contests in a row, but does not win five in a row is 0.0063
