Probability checkpoint 2

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. Never Former Current Non-daily Current Daily 0.576 0.215 0.04 0.169 Based on these data, what is the probability that a randomly selected U.S. adult currently smokes? A 0.209 B 0.169 C 0.04 D 0.0068 E none of these

A 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

Again, only 40% 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? A 0.24 B 0.48 C 0.36 D 0.16 E 0.52

E 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) = (0.40 * 0.40) + (0.60 * 0.60) = 0.16 + 0.36 = 0.52.

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. Here again is the information about the prescription 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 only a fever? A 0.24 B 0.27 C 0.30 D 0.36 E none of these

A 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).

Only 40% 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? A 0 B 0.16 C 0.80 D 0.64 E 0.25

B 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) = (.40) * (.40).

Here again is the information about the prescription 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 neither of the side effects? A 0.40 B 0.60 C 0.63 D 0.66 E 0.94

D 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.

Let A and B be two disjoint events such that P(A) = .20 and P(B) = .60. What is P(A and B)? A 0 B 0.12 C 0.68 D 0.80 E none of the above

A 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.

In a population, 3% have had a heart attack. Suppose a medical researcher randomly selects two people. Let X represent the event the first person has had a heart attack. Let Y represent the event the second person has had a heart attack. Which of the following is true about the two events? A X and Y are disjoint. B X and Y are independent. C None of the above are true. D Both (A) and (B) are true.

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

For safety reasons, four different alarm systems were installed in the vault containing the safety deposit boxes at a Beverly Hills bank. Each of the four systems will fail to detect theft with a probability of .01 independently of the others. The bank, obviously, is interested in the probability that when a theft occurs, at least one of the four systems will fail to detect it. This probability is equal to: A (.99)4 B 1 - (.99)4 C (.01)4 D 1 - (.01)4 E none of these

B We want to find the probability that at least one system fails to detect the theft. This is the complement of none fails to detect, which is the same as the complement of all detect. P(at least one fails to detect) = 1 - P(all detect) = 1 - (0.99)4.

For safety reasons, four different alarm systems were installed in the vault containing the safety deposit boxes at a Beverly Hills bank. Each of the four systems will fail to detect theft with a probability of .01 independently of the others. What is the probability that when a theft occurs, none of four systems will detect it? A (.99)4 B (.01)4 C (.01) * 4 D 4 * (.01) * (.99)3 E 4 * (.99) * (.01)3

B P(none detect) = P(all fail to detect) = P(1st fails to detect and 2nd fails to detect and 3rd fails to detect and 4th fails to detect). Since the systems work independently, we can multiply the individual probabilities. P(none detect) = P(all fail to detect) = P(1st fails to detect) * P(2nd fails to detect) * P(3rd fails to detect) * P(4th fails to detect) = (.01)4.

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? A 0.03 B 0.28 C 0.34 D 0.40 E 0.46

C We are given: P(S) = 0.10; P(F) = 0.30; P(S and F) = 0.06. We need to find P(S or F). Using the General Addition Rule: P(S or F) = P(S) + P(F) - P(S and F) = 0.10 + 0.30 - 0.06 = 0.34.

A fair die is rolled 12 times. Consider the following three possible outcomes: (i) 2 2 2 2 2 2 2 2 2 2 2 2 (ii) 1 1 2 2 3 3 4 4 5 5 6 6 (iii) 4 6 2 1 3 5 2 6 4 3 1 5 Which of the following is true? A It is absolutely impossible to get sequence (i). B (ii) is more likely than (i). C (iii) is more likely than (i) or (ii). D The three outcomes are equally likely. E Both (B) and (C) are true.

D 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).