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.
a) What is the probability that she has not distributed any toys by the end of her second visit?
b) What is the probability that she gives away the first toy on her fourth visit?
c) Given that she has given away her second toy on her fifth visit, what is the conditional probability that she will give away her third toy on her eighth visit?
d) What is the probability that she will give away the second toy on her fourth visit?
e) Given that she has not given away her second toy by her third visit, what is the conditional probability that she will give away her second toy on her fifth visit?
f) We will say that Marie “needs a new supply"" immediately after the visit on which she gives away her last toy. If she starts out with three toys, what is the probability that she completes at least five visits before she needs a new supply?
g) 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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a) To find the probability that she has not distributed any toys by the end of her second visit, we need to find the probability that the door is not answered and there is no toddler in residence for both her first and second visits.
The probability that the door is not answered on a visit is 1 - probability that the door is answered, which is 1 - (3/4) = 1/4.
The probability that there is no toddler in residence on a visit is 1 - probability that there is a toddler in residence, which is 1 - (1/3) = 2/3.
Since the events related to different households are independent, the probability that she has not distributed any toys by the end of her second visit is (1/4) * (2/3) = 1/6.
b) To find the probability that she gives away the first toy on her fourth visit, we need to find the probability that the first three visits result in no toy distribution and the fourth visit results in a toy distribution.
The probability that she does not distribute any toy on a visit is the same as the probability that she has not distributed any toys by the end of her second visit, which we found in part (a) to be 1/6.
The probability that she gives away a toy on a visit is 1 - probability that she does not distribute any toy, which is 1 - (1/6) = 5/6.
Since the events are independent, the probability that she gives away the first toy on her fourth visit is (1/6) * (1/6) * (1/6) * (5/6) = 5/1296.
c) Given that she has given away her second toy on her fifth visit, we need to find the conditional probability that she will give away her third toy on her eighth visit.
Since the events related to different households are independent, the probability that she gives away her third toy on her eighth visit is the same as the probability that she gives away her second toy on a visit, which we found in part (d) to be 5/1296.
d) To find the probability that she will give away the second toy on her fourth visit, we need to find the probability that the first three visits result in no toy distribution and the fourth visit results in a toy distribution.
We already found in part (b) that the probability of no toy distribution on a visit is 1/6, and the probability of a toy distribution on a visit is 5/6.
Hence, the probability that she gives away the second toy on her fourth visit is (1/6) * (1/6) * (1/6) * (5/6) = 5/1296.
e) Given that she has not given away her second toy by her third visit, we need to find the conditional probability that she will give away her second toy on her fifth visit.
The probability that she has not given away the second toy by her third visit is the same as the probability that she has not distributed any toys by the end of her second visit, which we found in part (a) to be 1/6.
Since the events are independent, the probability that she gives away her second toy on her fifth visit is the same as the probability that she gives away her first toy on a visit, which we found in part (b) to be 5/1296.
f) To find the probability that Marie completes at least five visits before she needs a new supply, we need to consider the following possibilities:
1. She distributes all three toys in the first four visits, which has probability (5/6) * (5/6) * (5/6) = 125/216.
2. She distributes two toys in the first four visits and then needs a new supply. The probability of this happening is (5/6) * (5/6) * (1/6) * (5/6) = 125/1296.
3. She distributes one toy in the first four visits and then needs a new supply. The probability of this happening is (5/6) * (1/6) * (5/6) * (5/6) = 125/1296.
4. She distributes no toys in the first four visits and then needs a new supply. The probability of this happening is (1/6) * (5/6) * (5/6) * (5/6) = 125/1296.
The probability that she completes at least five visits before she needs a new supply is the sum of these probabilities: (125/216) + (125/1296) + (125/1296) + (125/1296) = 725/1296.
g) To find the expected value of the number of houses with toddlers that Marie visits without leaving any toys before she needs a new supply, we need to consider the possible scenarios:
1. She distributes all six toys before needing a new supply, which means she visits six houses with toddlers without leaving any toys. The probability of this happening is (5/6)^6 = 15625/46656.
2. She distributes five toys before needing a new supply, which means she visits five houses without leaving any toys. The probability of this happening is (5/6)^5 * (1/6) * (1/6) * (1/6) * (1/6) = 625/7776.
3. She distributes four toys before needing a new supply, which means she visits four houses without leaving any toys. The probability of this happening is (5/6)^4 * (1/6)^2 * (1/6) = 125/1296.
The expected value is calculated by multiplying each scenario by its respective probability and summing them up:
(6 * (15625/46656)) + (5 * (625/7776)) + (4 * (125/1296)) = 670/2187.