Marie distributes toys for toddlers. She makes visits to households and gives away one toy only on visits for which the door is answered and a toddler is in residence. On any visit, the probability of the door being answered is 3/4, and the probability that there is a toddler in residence is 1/3. Assume that the events “Door answered" and “Toddler in residence" are independent and also that events related to different households are independent.
1.What is the probability that she has not distributed any toys by the end of her second visit?
2.What is the probability that she has not distributed any toys by the end of her second visit?
3.What is the probability that she has not distributed any toys by the end of her second visit?
4.What is the probability that she will give away the second toy on her fourth visit?
5.What is the probability that she will give away the second toy on her fourth visit?
6.What is the probability that she will give away the second toy on her fourth visit?
7.If she starts out with exactly six toys, what is the expected value of the number of houses with toddlers that Marie visits without leaving any toys (because the door was not answered) before she needs a new supply?
1. The probability that she has not distributed any toys by the end of her second visit is the probability that the door is not answered and there is no toddler in residence on both visits. Since these events are independent, the probability is (1/4) * (2/3) = 1/6.
2. The probability that she has not distributed any toys by the end of her second visit is the same as the probability in question 1, which is 1/6.
3. The probability that she has not distributed any toys by the end of her second visit is the same as the probability in question 1, which is 1/6.
4. The probability that she will give away the second toy on her fourth visit is the probability that the door is answered and there is a toddler in residence on the first three visits, and the door is not answered or there is no toddler in residence on the fourth visit. Since these events are independent, the probability is (3/4) * (1/3) * (3/4) * (2/3) * (1/4) = 1/16.
5. The probability that she will give away the second toy on her fourth visit is the same as the probability in question 4, which is 1/16.
6. The probability that she will give away the second toy on her fourth visit is the same as the probability in question 4, which is 1/16.
7. The expected value of the number of houses with toddlers that Marie visits without leaving any toys before she needs a new supply is equal to the expected number of visits without leaving any toys multiplied by the probability of there being a toddler in residence. The probability of there being a toddler in residence is 1/3.
To find the expected number of visits without leaving any toys, we can think of it as a geometric distribution, where the probability of success (leaving a toy) is 1/4. The expected number of visits without leaving any toys is equal to 1 divided by the probability of success, which is 1 divided by (1/4) = 4.
Therefore, the expected value of the number of houses with toddlers that Marie visits without leaving any toys before she needs a new supply is (4) * (1/3) = 4/3.
1. To find the probability that Marie has not distributed any toys by the end of her second visit, we need to consider the complementary event—the event that she has distributed at least one toy.
The probability of a toddler being in residence is 1/3, and the probability of the door being answered is 3/4. So, the probability of her giving away a toy on a visit is (1/3) * (3/4) = 1/4.
Therefore, the probability of not giving away a toy on a visit is 1 - 1/4 = 3/4.
Since each visit is independent, the probability of her not giving away a toy on both her first and second visits is (3/4) * (3/4) = 9/16.
So, the probability that she has not distributed any toys by the end of her second visit is 9/16.
2. The probability that Marie has not distributed any toys by the end of her second visit is 9/16.
3. The probability that Marie has not distributed any toys by the end of her second visit is again 9/16.
4. To find the probability that Marie will give away the second toy on her fourth visit, we need to consider the probability of her not giving away any toy on her previous three visits.
The probability of not giving away a toy on a visit is 3/4 (as calculated in question 1). Since each visit is independent, the probability of her not giving away any toy on three visits (first, second, and third) is (3/4) * (3/4) * (3/4) = 27/64.
Therefore, the probability of her giving away the second toy on her fourth visit is 1 - 27/64 = 37/64.
5. The probability that Marie will give away the second toy on her fourth visit is 37/64.
6. The probability that Marie will give away the second toy on her fourth visit remains 37/64.
7. If Marie starts out with exactly six toys, she will need to replenish her supply when she runs out of toys. We need to find the expected value of the number of houses with toddlers that she visits without leaving any toys (because the door was not answered) before needing a new supply.
The probability of a toddler being in residence is 1/3, and the probability of the door being answered is 3/4. So, the probability of her not giving away a toy on a visit is 1 - (1/3) * (3/4) = 2/3.
Let's assume she visits x houses without giving away any toys before needing a new supply. The probability of this event happening is (2/3)^x.
The expected value of x can be calculated using the formula E(x) = ∑(x * P(x)), where P(x) represents the probability of x happening.
Since the number of houses Marie visits without leaving any toys before needing a new supply follows a geometric distribution, the expected value can be calculated as E(x) = (1 - P(x)) / P(x).
In this case, P(x) = 2/3, and we want to find E(x) when she starts out with six toys.
E(x) = (1 - (2/3)) / (2/3) = 3/2.
Therefore, the expected value of the number of houses with toddlers that Marie visits without leaving any toys before needing a new supply is 3/2.