Probability Checkpoint 2

Let A and B be two disjoint events such that P(A) = .20 and P(B) = .60. What is P(A and B)?

If two events are disjoint, then by definition, the two events cannot happen together. The probability of these two events happening together is denoted by P(A and B). If this is impossible, then P(A and B) = 0.

Only 30% of the students in a certain liberal arts college are males. If two students from this college are selected at random, what is the probability that they are both males?

Let M1 = the first person is a male. Let M2 = second person is a male. We want P(M1 and M2). Because the population is fairly large, the events are independent, and we can use the Multiplication Rule for Independent Events. Therefore, P(M1 and M2) = P(M1) * P(M2) = (.30) * (.30).

Again, only 30% of the students in a certain liberal arts college are males. If two students from this college are selected at random, what is the probability that they are of the same gender?

P(both of the same gender) = P(2 males or 2 females). Since these are disjoint events: P(2 males or 2 females) = P(2 males) + P(2 females). Since the choices are independent: P(2 males) + P(2 females) = (0.30 * 0.30) + (0.70 * 0.70) = 0.09 + 0.49 = 0.58.

A particular student has a lot of trouble getting up in the morning. To make sure he will not oversleep, he sets four identical alarm clocks. Each of the four alarms will buzz at the set time with a probability of .98 independently of the others. What is the probability that when the set time arrives, all four alarms will buzz?

Remember that, by definition, a probability must be between 0 and 1, inclusive. We want to find P(all buzz) = P(1st buzzes and 2nd buzzes and 3rd buzzes and 4th buzzes). Since the alarms work independently, we can multiply the individual probabilities. Correct Answer is : (.98)4

A fair die is rolled 12 times. Consider the following three possible outcomes: (i) 6 5 4 3 2 1 6 5 4 3 2 1 (ii) 1 1 1 1 1 1 1 1 1 1 1 1 (iii) 3 6 2 1 5 4 2 5 1 6 4 3 Which of the following is true?

The die is fair. This means that all faces have an equal probability of occurring on any given roll (1/6). Since each roll is independent of the other rolls, the probability of the each of the three sequences shown is the same, (1/6)12. So the three sequences are equally likely (or we could say equally unlikely since each has such a small chance of occurring).

In a population, 5% of the females have had a kidney stone. Suppose a medical researcher randomly selects two females. Let X represent the event the first female has had a kidney stone. Let Y represent the event the second female has had a kidney stone. Which of the following is true about the two events?

The occurrence of X does not affect the probability of Y since the women are randomly selected from a large population. So the events are independent. Also, these two events overlap. If K = had a kidney stone and N = no kidney stone, then KK is a possible outcome. Thus, the two events are not disjoint. Besides, recall that if events are independent, they cannot be disjoint.X and Y are independent.

There is a 10% chance of experiencing a sore throat (S) and a 30% chance of experiencing a fever (F). The information also states that there is a 6% chance of experiencing both side effects. What is the probability of experiencing only a fever?

The probability of experiencing only a fever is P(F) - P(S and F) = 0.30 - 0.06 = 0.24. You could also build a probability table to find P(F and not S).

The following probabilities are based on data collected from U.S. adults during the National Health Interview Survey 2005-2007. Individuals are placed into a smoking category based on whether they ever smoked 100 cigarettes (in their lifetime) and their behavior in the last 30 days.Based on these data, what is the probability that a randomly selected U.S. adult currently smokes?

We want to find P( current non-daily or current daily ). Since these are disjoint events, we can add the two probabilities. 0.04 + 0.169 = 0.209

According to the information that comes with a certain prescription drug, when taking this drug, there is a 10% chance of experiencing a sore throat (S) and a 30% chance of experiencing a fever (F). The information also states that there is a 6% chance of experiencing both side effects. What is the probability of experiencing a sore throat or a fever?

We want to find P(S or F), which cannot be 0.28, because it has to be greater than either individual probability. You may have made a probability table and added P(S and not F) to P(not S and F). This is incorrect because P(S or F) includes the situation where both a sore throat and a fever occur. 0.34 is the correct answer.

A particular student has a lot of trouble getting up in the morning. To make sure he will not oversleep, he sets four identical alarm clocks. Each of the four alarms will buzz at the set time with a probability of .98 independently of the others. The student, obviously, is interested in the probability that when the set time occurs, at least one of the four alarms will buzz. This probability is equal to:

We want to find the probability that at least one alarm buzzes at the set time, which is the complement of none buzzes. P(at least one buzzes) = 1 - P(none buzzes) = 1 - (0.02)4.

There is a 10% chance of experiencing a sore throat (S) and a 30% chance of experiencing a fever (F). The information also states that there is a 6% chance of experiencing both side effects. What is the probability of experiencing neither of the side effects?

You want to compute P(neither side effect) = P(not S and not F) = P(not S or F) = 1 - 0.34 = 0.66. You may find that using a probability table is simpler than using formulas.