Online Psych Stats

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:

1 - (.02)4

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?

X and Y are independent

Of the at-home games, we are interested in finding what proportion were wins. In order to figure this out, we need to find:

Good job! We are limiting our interest to the home games only, so we condition on H. Out of the home games, we are interested in the probability of winning (W), therefore we are interested in P(W | H).

Again here is the information about the characteristics of a basketball team's season: 60% of all the games were at-home games. Denote this by H (the remaining were away games). 35% of all games were wins. Denote this by W (the remaining were losses). 25% of all games were at-home wins. Of the at-home games, what proportion of games were wins?

Good job! We want to find P(W | H). We apply the definition of conditional probability: P(W | H) = P(H and W) / P(H) = .25 / .60 ≈ .42

Let A and B be two disjoint events such that P(A) = .20 and P(B) = .70. 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.

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?

P(neither side effect) = P('not S'and 'not F') = P(not'S or F') = 1 - 0.34 = 0.66.

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

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?

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.

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 Probability 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?

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