3.2b Mutually exclusive events

A couple has two children. Let A be the event that their first child is a boy, and note that P(A)=51.2%. Let B be the event that their second child is a girl, with P(B)=48.8%. A and B are independent events. What is the probability that the couple has a first child that is a boy and a second child that is a girl? Give your answer as a decimal rounded to two decimal places.

0.25 Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=51.2%⋅48.8%=0.512⋅0.488≈0.25

An electric company has to track on-time payments and late payments for each customer monthly. It is impossible for a customer to pay on-time and late each month. If the probability that a customer pays on-time each month is 0.55, and the probability that a customer pays late or on-time each month is 0.82, what is the probability that a customer pays late each month?

0.27 Because it is impossible for a customer to pay on-time and late each month, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.82−0.55=0.27 The probability that a customer pays late each month is 0.27.

If a police officer pulls over someone for speeding, the police officer can either give a ticket or a warning, so it is impossible for a police officer to give a ticket and a warning for speeding. If the probability that a police officer will give a warning for speeding is 0.03, and the probability that a police officer will give a ticket or a warning for speeding is 0.52, what is the probability that a police officer will give a ticket for speeding?

0.49 Because it is impossible for a police officer to give a ticket and a warning for speeding, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find that P(A)=P(A OR B)−P(B) So plugging in the given values, we find P(A)=0.52−0.03=0.49 The probability that a police officer will give a ticket for speeding is 0.49.

Suppose A and B are mutually exclusive events, and that P(B)=0.48 and P(A OR B)=0.98. Find P(A).

0.50 Remember that when A and B are mutually exclusive events, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find that P(A)=P(A OR B)−P(B) So plugging in the given values, we find P(A)=0.98−0.48=0.5

If A and B are independent events with P(A)=0.90 and P(A AND B)=0.54, find P(B). Give your answer as a decimal rounded to two decimal places.

0.60 Remember that for independent events, P(A AND B)=P(A)P(B) Solving for P(B), we find that P(B) = P(A and B)/P(A) P(B)=0.54/0.90 = 0.60

Randy wants to either ride share to work or drive his own car to work, but it is impossible for Randy to ride share and drive his own car in one trip. If the probability that Randy ride shares is 0.22, and the probability that Randy drives his own car is 0.42, what is the probability that Randy ride shares or drives his own car to work?

0.64 Because it is impossible for Randy to ride share and drive his own car in one trip, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) So we find that P(A OR B)=P(A)+P(B)=0.22+0.42=0.64 The probability that Randy ride shares or drives his own car to work is 0.64.

Consider the Venn diagram below. Each of the dots outside the circle represents a graduating college student surveyed about their post-college job search. Five of the graduates are business majors: represented by red dots and labeled: 1,2,3,4,5 (abbreviated BM); Four of the graduates are social work majors: and are represented by blue dots: 1,2,3,4 (abbreviated SW). The two events represented in the Venn diagram are: Event A: The student has applied to at least one post-college job. Event B: The student currently is working in an internship position. Three of the business majors BM1,BM2,BM4 and two of the social work majors SW2,SW3 say they have applied to at least one job. Business majors BM2,BM3 state that they are currently working in an internship position. Social work majors SW2,SW4 say they are also currently working in an internship position. Students BM5,SW1 say that neither of the options currently apply to them. Place the dots in the appropriate event given the information above (you may not use all of the dots), then determine if the events are mutually exclusive.

A and B are not mutually exclusive. We say that two events are mutually exclusive if they share no outcomes, that is, P(A and B)=0. Here, A and B share an outcome: BM2,SW2. Thus, P(A and B)≠0 and the events are not mutually exclusive.

You have a standard deck of 52 cards. There are four suits: clubs, diamonds, hearts, and spades. Each suit has 13 cards: Ace,2,3,4,5,6,7,8,9,10,Jack,Queen,King. Given Events A and B, are the two events mutually exclusive? Explain your answer. Event A: Drawing a 10.

No, the events are not mutually exclusive because they share the common outcome of 10 of hearts. A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B)=0. In this case, A and B have an outcome in common, 10 of hearts, so they are not mutually exclusive.

If you roll a fair 6-sided die and then roll a second fair die, what is the probability that you roll a 1, 2, or 3 on the first die and roll a 5 on the second die?

Note that these are independent events because the outcome when you roll a fair die does not affect the outcome when you roll a second fair die. So by the multiplication rule for independent events, we can take the probability of each event and multiply them.The probability that you roll a 1, 2, or 3 on the first die is 3/6, and the probability that you roll a 5 on the second die is 1/6. The probability that you roll a 1, 2, or 3 on the first die and roll a 5 on the second die is 3/6⋅1/6=1/12

At a major international airport, passengers are questioned about their destination. Given Events A and B, are the two events mutually exclusive? Explain your answer. Event A: The passengers are traveling to Paris, France. Event B: The passengers are NOT traveling to Paris, France. Select the correct answer below:

Yes, the events are mutually exclusive because they have no outcomes in common. A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B)=0. In this case, A and B do not have an outcome in common, so they are mutually exclusive.

Two friends are both pregnant, and find out they are each expecting twins! Let A be the event that one friend is pregnant with identical twins, and note that P(A)=0.0045. Let B be the event that the other friend is pregnant with fraternal twins, and note that P(B)=0.01. A and B are independent events. What is the probability that one friend is pregnant with identical twins, and one friend is pregnant with fraternal twins? Give your answer as a percent, rounded to four decimal places if necessary.

.0045% Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=0.0045⋅0.01=0.000045 To express our answer as a percent, we multiply by 100: 0.000045×100=0.0045%

Pregnant women have the option of being scanned for Cystic Fibrosis risks in their unborn babies. If a mother or a father have a certain recessive gene, the baby is at risk for Cystic Fibrosis. Given the three events, which of the following statements is true? Select all that apply. Event A: The mother or father carries the recessive gene. Event B: The father carries the recessive gene. Event C: The baby is at risk for Cystic Fibrosis.

Event A and Event C are not mutually exclusive. Remember that in general, A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B)=0. In this question, note that if the mother or father carries the recessive gene for cystic fibrosis, the baby is at risk. So here, A and C are not mutually exclusive, and B and C are not mutually exclusive. A and B are not mutually exclusive because they share an outcome: the father carrying the recessive gene.