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.
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?
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