Problem 1. Marie gives away children toys
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.
What is the probability that she has not distributed any toys by the end of her second visit?
9/16
What is the probability that she gives away the first toy on her fourth visit?
(27/64)*(1/4)
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?
unanswered
What is the probability that she will give away the second toy on her fourth visit?
unanswered
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?
unanswered
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?
unanswered
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?
unanswered
Could any nice guy help me with the answers? Thanks in advance.
1. 9/16
2. 24/256
3. 9/64
4. 27/256
5. 1/8
6. 243/256
7. 2
2. 27/256
Sure! I can help you with some of the answers.
To find the probability that Marie has not distributed any toys by the end of her second visit, we can use the concept of independent events. Since the events of the door being answered and a toddler being in residence are independent, the probability that both events do not occur on each visit is (1-3/4)*(1-1/3) = 1/4. Since we are looking at the first two visits, the probability is (1/4)*(1/4) = 1/16. Therefore, the probability that Marie has not distributed any toys by the end of her second visit is 1 - 1/16 = 15/16.
To find the probability that she gives away the first toy on her fourth visit, we can use the same concept of independent events. The probability that the door is answered on the first three visits without a toddler in residence is (3/4)*(2/3)*(2/3) = 4/9. The probability that the door is answered on the fourth visit with a toddler in residence is (3/4)*(1/3) = 1/4. Therefore, the probability that she gives away the first toy on her fourth visit is (4/9)*(1/4) = 1/9.
I hope these answers help you. If you have any more questions, feel free to ask!
Sure! Let's go through each question one by one:
1. What is the probability that Marie has not distributed any toys by the end of her second visit?
To solve this, we need to determine the probability that Marie does not give away any toys on both her first and second visits. Since the events are independent, we can multiply the probabilities together.
P(no toys on first visit) = P(door not answered) * P(no toddler in residence) = (1 - P(door answered)) * (1 - P(toddler in residence))
P(no toys on second visit) = P(door not answered) * P(no toddler in residence) = (1 - P(door answered)) * (1 - P(toddler in residence))
To find the probability that both events occur, we multiply the probabilities together:
P(no toys on both visits) = P(no toys on first visit) * P(no toys on second visit)
Substituting the given probabilities (P(door answered) = 3/4 and P(toddler in residence) = 1/3):
P(no toys on both visits) = (1 - 3/4) * (1 - 1/3) = 1/4 * 2/3 = 2/12 = 1/6 = 1/6
So, the probability that she has not distributed any toys by the end of her second visit is 1/6.
2. What is the probability that she gives away the first toy on her fourth visit?
To solve this, we need to determine the probability that Marie does not give away any toys on her first three visits (3/4 probability each time), and then on her fourth visit, the door is answered (1/4 probability) and there is a toddler in residence (1/3 probability).
P(no toys on first three visits) = (3/4) * (3/4) * (3/4)
P(toy on fourth visit) = (1/4) * (1/3)
To find the probability that both events occur, we multiply the probabilities together:
P(no toys on first three visits and toy on fourth visit) = P(no toys on first three visits) * P(toy on fourth visit)
Substituting the given probabilities:
P(no toys on first three visits and toy on fourth visit) = (3/4) * (3/4) * (3/4) * (1/4) * (1/3) = (27/256) * (1/4) * (1/3) = 27/3072
So, the probability that she gives away the first toy on her fourth visit is 27/3072.
3. 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?
To answer this question, we need to use conditional probability. If we are given that Marie has already given away her second toy on her fifth visit, we can ignore the previous events and focus on the new scenario.
The conditional probability is given by:
P(third toy on eighth visit | second toy on fifth visit) = P(third toy on eighth visit and second toy on fifth visit) / P(second toy on fifth visit)
Since the events related to different households are independent, we can calculate each probability separately:
P(third toy on eighth visit and second toy on fifth visit) = (3/4) * (3/4) * (1/4) * (1/3) * (1/4) * (1/3) * (1/4) * (1/3)
P(second toy on fifth visit) = (3/4) * (3/4) * (3/4) * (3/4) * (1/4) * (1/3) * (1/4) * (1/3)
Substituting the given probabilities:
P(third toy on eighth visit | second toy on fifth visit) = [(3/4) * (3/4) * (1/4) * (1/3) * (1/4) * (1/3) * (1/4) * (1/3)] / [(3/4) * (3/4) * (3/4) * (3/4) * (1/4) * (1/3) * (1/4) * (1/3)]
Simplifying the expression, we get:
P(third toy on eighth visit | second toy on fifth visit) = 1/3
So, the conditional probability that she will give away her third toy on her eighth visit, given she has given away her second toy on her fifth visit, is 1/3.
I hope this helps! Let me know if you have any further questions.