Question

Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome.

a. pF can take any value between 0 and [infinity]. qF can take any value from 0 to [infinity].
b. pF can only take values 1, 2, 3, 4, or 5. qF can only take values 0 or 1.
c. pF can take values 1, 2, 3, 4, or 5. qF can take any value from 0 to [infinity].
d. pF can take any value between 0 and [infinity]. qF can only take values 0 or 1.

Answer 1

The values of c) pF can take values 1, 2, 3, 4, or 5, while qF can take any value from 0 to infinity. The probability of the first occurrence of event F being on the second trial is 2/9.

Step-by-step explanation:

The event F is defined as rolling a four or a five. In this case, pF represents the number of times you need to roll the die to obtain the first four or five as the outcome, and qF represents the number of times you do not roll a four or five before obtaining it.

Since you are interested in how many times you need to roll the die to obtain the first four or five, pF can take values 1, 2, 3, 4, or 5, while qF can take any value from 0 to infinity. Therefore, option c is correct.

To find the probability that the first occurrence of event F is on the second trial, we can calculate the probability of not rolling a four or five on the first trial and then rolling a four or five on the second trial.

The probability of not rolling a four or five on the first trial is qF = 2/3, since there are four outcomes that are not a four or five out of six possible outcomes.

The probability of rolling a four or five on the second trial is pF = 2/6, since there are two outcomes that are a four or five out of six possible outcomes.

Therefore, the probability that the first occurrence of event F is on the second trial is pF * qF = (2/3) * (2/6) = 4/18

= 2/9.