Fred is giving away samples of dog food. He makes visits door to door, but he gives a sample away (one can of dog food) only on those visits for which the door is answered and a dog is in residence. On any visit, the probability of the door being answered is 3/4 , and the probability that any given household has a dog is 2/3 .
Assume that the events “Door answered" and “A dog lives here" are independent and also that events related to different households are independent.
1. What is the probability that Fred gives away his first sample on his third visit?
2. Given that he has given away exactly four samples on his first eight visits, what is the conditional probability that Fred will give away his fifth sample on his eleventh visit?
3. What is the probability that he gives away his second sample on his fifth visit?
4. Given that he did not give away his second sample on his second visit, what is the conditional probability that Fred will give away his second sample on his fifth visit?
5. We will say that Fred “needs a new supply" immediately after the visit on which he gives away his last sample. If he starts out with two samples, what is the probability that he completes at least five visits before he needs a new supply?
6. If he starts out with exactly 10 samples, what is the expected value of the number of homes with dogs where Fred visits but leaves no samples (because the door was not answered) before he needs a new supply?
To solve these problems, we will use the concepts of probability and independence. Let's solve each question step by step:
1. To find the probability that Fred gives away his first sample on his third visit, we need to consider two independent events: the door being answered and a dog living at the residence. The probability of the door being answered is 3/4, and the probability that a household has a dog is 2/3. Therefore, the probability that Fred gives away his first sample on his third visit is:
P(Door answered) * P(Dog lives here) = (3/4) * (2/3) = 1/2
2. Given that Fred has given away exactly four samples on his first eight visits, we know that there are four successful visits (door answered, dog present) and four unsuccessful visits (either door not answered or no dog present). Now, we want to find the conditional probability that Fred gives away his fifth sample on his eleventh visit.
In this case, we have four successful visits in the first eight visits. Therefore, the probability of a successful visit is 4/8 = 1/2. Since the events are independent, the probability of a successful visit on the eleventh visit is also 1/2.
Therefore, the conditional probability that Fred gives away his fifth sample on his eleventh visit is 1/2.
3. To find the probability that Fred gives away his second sample on his fifth visit, we need to consider the same independent events as in question 1. The probability of the door being answered is 3/4, and the probability that a household has a dog is 2/3.
Therefore, the probability that Fred gives away his second sample on his fifth visit is:
P(Door answered) * P(Dog lives here) = (3/4) * (2/3) = 1/2
4. Given that Fred did not give away his second sample on his second visit, we want to find the conditional probability that Fred gives away his second sample on his fifth visit.
Since Fred did not give away his second sample on his second visit, it means that his second visit was unsuccessful (either the door was not answered or no dog was present). So, we need to find the conditional probability that Fred gives away his second sample on his fifth visit, given that his second visit was unsuccessful.
Since the events related to different households are independent, the probability of a successful visit on the fifth visit is the same as in question 3, which is 1/2.
Therefore, the conditional probability that Fred gives away his second sample on his fifth visit, given that his second visit was unsuccessful, is 1/2.
5. If Fred starts out with two samples, the maximum number of successful visits he can have is two because he only has two samples to give away. Therefore, he needs at least three more visits before he needs a new supply.
Now, we need to find the probability that Fred completes at least five visits before he needs a new supply. Let's break down this probability:
P(Completes at least five visits) = P(Completes five visits) + P(Completes six visits) + P(Completes seven visits) + P(Completes eight visits) + ...
Since the events related to different households are independent, the probability of completing a specific number of visits is the product of the probability of a successful visit (door answered, dog present) and the probability of an unsuccessful visit (either door not answered or no dog present) raised to the power of the number of unsuccessful visits.
P(Completes at least five visits) = (1/2)^3 + (1/2)^4 + (1/2)^5 + ... (infinite sum)
This is an infinite geometric series with a common ratio of (1/2). Using the formula for the sum of an infinite geometric series, we get:
P(Completes at least five visits) = (1/2)^3 / (1 - 1/2) = (1/2)^3 / (1/2) = 1/8 / 1/2 = 1/4
Therefore, the probability that Fred completes at least five visits before he needs a new supply is 1/4.
6. If Fred starts out with exactly ten samples, he needs to continue giving away samples until he runs out. The expected value of the number of homes with dogs where Fred visits but leaves no samples can be calculated using the concept of expected value.
We need to find the expected number of visits where the door is answered but no sample is given away for each household. The probability of a visit with no sample given away is (1 - P(Dog lives here)). Using the given probability that any given household has a dog is 2/3, we have:
Expected number of visits without sample given away = 10 * (1 - 2/3) = 10 * (1/3) = 10/3
Therefore, the expected value of the number of homes with dogs where Fred visits but leaves no samples before he needs a new supply is 10/3.