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?
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What is the probability that she gives away the first toy on her fourth visit?
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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?
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What is the probability that she will give away the second toy on her fourth visit?
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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?
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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?
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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?
To answer these questions, we need to understand the basic principles of probability and apply them to the given scenario. Let's go question by question:
1. What is the probability that she has not distributed any toys by the end of her second visit?
To find this probability, we need to calculate the probability of both the first and second visits having the door unanswered or no toddlers in residence. The probability of the door being unanswered on each visit is 1 - 3/4 = 1/4, and the probability of no toddlers being in residence is 1 - 1/3 = 2/3. Since the visits are independent, the probability that both visits have no distribution is (1/4) * (2/3) = 1/6. Therefore, 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?
In order for Marie to give away her first toy on her fourth visit, the first three visits must have either the door unanswered or no toddlers in residence, and the fourth visit must have both the door answered and a toddler in residence. The probability of the door being unanswered on each of the first three visits is 1/4, and the probability of no toddlers in residence is 2/3. The probability of the door being answered and a toddler being in residence on the fourth visit is (3/4) * (1/3) = 1/4. Since the visits are independent, the probability of this sequence of events happening is (1/4) * (1/4) * (1/4) * (1/4) = 1/256.
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?
Here, we need to find the conditional probability that she will give away her third toy on her eighth visit, given that she has already given away her second toy on her fifth visit.
Since the events related to different households are independent, the number of visits for giving away toys on each household is independent. Therefore, the conditional probability of giving away the third toy on the eighth visit is the same as the unconditional probability of giving away the third toy on any visit.
4. What is the probability that she will give away the second toy on her fourth visit?
Similar to question 2, in order for Marie to give away her second toy on her fourth visit, the first three visits must have either the door unanswered or no toddlers in residence, and the fourth visit must have both the door answered and a toddler in residence. The probability of this sequence of events happening is (1/4) * (1/4) * (1/4) * (1/4) = 1/256.
5. 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?
To find this conditional probability, we need to divide the probability that she gives away her second toy on her fifth visit and has not given it away on her third visit by the probability that she has not given away her second toy by her third visit.
The probability that she gives away her second toy on her fifth visit is calculated in the same way as in question 4, which is 1/256. The probability that she has not given away her second toy on her third visit is (1 - (1/4) * (2/3)) = 7/12.
So, the conditional probability in this case would be (1/256) / (7/12).
6. If she starts out with three toys, what is the probability that she completes at least five visits before she needs a new supply?
Let's consider the case where she completes exactly four visits without needing a new supply. For this to happen, she must give away toys on the first four visits, and then not give any toys on the fifth visit. The probability of giving away a toy on each visit is (3/4) * (1/3) = 1/4. Therefore, the probability of completing exactly four visits without needing a new supply is (1/4)^4 = 1/256.
The probability of completing at least five visits before needing a new supply would be 1 - the probability of completing exactly four visits without needing a new supply, which is 1 - 1/256 = 255/256.
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 before she needs a new supply?
To find the expected value, we need to calculate the probability of visiting each possible number of houses without leaving any toys and multiply it by the corresponding number of houses. Since Marie starts with six toys, the maximum number of houses she can visit without needing a new supply is six.
The expected value is calculated as follows: (0)*(probability of visiting 0 houses) + (1)*(probability of visiting 1 house) + (2)*(probability of visiting 2 houses) + ... + (6)*(probability of visiting 6 houses).
The probability of visiting each number of houses without leaving any toys can be calculated using the binomial distribution formula. For example, the probability of visiting exactly zero houses without leaving any toys is calculated as (1/4)^0 * (3/4)^6.
By summing up the probabilities multiplied by the corresponding number of houses for all possible numbers of houses, we can find the expected value.