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.

What is the probability that Fred gives away his first sample on his third visit?

- unanswered

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?

- unanswered

What is the probability that he gives away his second sample on his fifth visit?

- unanswered

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?

- unanswered

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?

- unanswered

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?
- unanswered

1)17/10

2)41/17
3)b/a

sorry guys here are not the correct ans for this problem.

CORRECT ANSWERS :

1)1/8
2)1/8
3)1/8

1)1/8

2)1/8
3)1/8
4)1/6
5)5/16
6)10/3

CORRECT ANSWERS

To solve these probability problems, we will use the concepts of independence and the multiplication rule of probability.

1. Probability of giving away the first sample on the third visit:
Since we know the probability of the door being answered is 3/4 and the probability of a dog living in a household is 2/3, we can multiply these probabilities together to get the probability of both events occurring. However, we need the door to be answered and a dog to be present for Fred to give away a sample. Thus, the probability is:
P(Door answered and Dog present) = P(Door answered) * P(Dog present) = (3/4) * (2/3) = 1/2.

2. Conditional probability of giving away the fifth sample on the eleventh visit, given that four samples were given away on the first eight visits:
To calculate this conditional probability, we need to consider the previous visits and the given information. We know that four samples were given away in the first eight visits, so there are four visits remaining. The probability of giving away the fifth sample on the eleventh visit is then:
P(Fifth sample on eleventh visit | four samples in first eight visits) = P(Fifth sample on eleventh visit and four samples in first eight visits) / P(Four samples in first eight visits).

To calculate the numerator, we use the multiplication rule of probability. We need to find the probability of four samples given away in the first eight visits and the fifth sample given away on the eleventh visit. Since the events related to different households are independent, we can calculate it as:
P(Fifth sample on eleventh visit and four samples in first eight visits) = P(Four samples in first eight visits) * P(Fifth sample on eleventh visit | four samples in first eight visits) = (3/4)^4 * (1/4).

To calculate the denominator, we consider the probability of four samples given away in the first eight visits, regardless of when the fifth sample is given away. Again, the events are independent, so we have:
P(Four samples in first eight visits) = (3/4)^4 * (1/4) + (1/4)^4 * (3/4) + 4 * (3/4)^4 * (1/4) * (1/4)^3.

Finally, we can calculate the conditional probability:
P(Fifth sample on eleventh visit | four samples in first eight visits) = (P(Fifth sample on eleventh visit and four samples in first eight visits)) / P(Four samples in first eight visits).

3. Probability of giving away the second sample on the fifth visit:
Using the same reasoning as in question 1, we can calculate the probability as:
P(Door answered and Dog present) = P(Door answered) * P(Dog present) = (3/4) * (2/3) = 1/2.

4. Conditional probability of giving away the second sample on the fifth visit, given that the second sample was not given away on the second visit:
Similar to question 2, we need to consider the previous visits and the given information. We know that the second sample was not given away on the second visit, so there is one less visit remaining. Thus, the conditional probability is:
P(Second sample on fifth visit | second sample not given away on second visit) = P(Second sample on fifth visit and second sample not given away on second visit) / P(Second sample not given away on second visit).

To calculate the numerator, we can use the multiplication rule of probability. We need to find the probability of the second sample not given away on the second visit and the second sample given away on the fifth visit. Since the events related to different households are independent, we can calculate it as:
P(Second sample on fifth visit and second sample not given away on second visit) = P(Second sample not given away on second visit) * P(Second sample on fifth visit | second sample not given away on second visit) = (1/4) * (3/4).

To calculate the denominator, we consider the probability of the second sample not given away on the second visit, regardless of when the second sample is given away. Again, the events are independent, so we have:
P(Second sample not given away on second visit) = (1/4) * (3/4) + (3/4) * (1/4) + 2 * (1/4) * (3/4)^2 * (1/4).

Finally, we can calculate the conditional probability:
P(Second sample on fifth visit | second sample not given away on second visit) = (P(Second sample on fifth visit and second sample not given away on second visit)) / P(Second sample not given away on second visit).

5. Probability of completing at least five visits before needing a new supply, starting with two samples:
This can be interpreted as the complement of the probability of needing a new supply within the first four visits. Since Fred gives away a sample only when the door is answered and a dog is present, the probability of not needing a new supply within the first four visits is the probability that either the door is unanswered or no dog is present in all four visits.

The probability that the door is unanswered in a visit is 1 - (3/4) = 1/4, and the probability that no dog is present in a visit is 1 - (2/3) = 1/3. Since events related to different households are independent, we can multiply these probabilities to find the probability that the door is unanswered and no dog is present for each of the four visits.

P(No new supply within the first four visits) = (1/4)^4 * (1/3)^4.

Finally, we can calculate the probability of completing at least five visits before needing a new supply as the complement of the probability above:
P(Completing at least five visits before needing a new supply) = 1 - P(No new supply within the first four visits).

6. Expected value of the number of homes with dogs where Fred visits but leaves no samples before needing a new supply, starting with ten samples:
To calculate the expected value, we need to consider the probabilities of different outcomes and their corresponding values. In this case, we are interested in the number of homes with dogs visited by Fred, where no samples are given away because the door was not answered.

Since the door is answered with a probability of 3/4, the door is unanswered with a probability of 1 - (3/4) = 1/4. We are interested in the number of homes with dogs, so we multiply this probability by the expectation of the number of homes with dogs visited by Fred, which is 10 (initial number of samples).

Expected value = (1/4) * 10 = 5.

Therefore, we expect Fred to visit and leave no samples at five homes with dogs before needing a new supply if he starts with exactly ten samples.