Problem description
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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.
find out the probability for below 2 scenarios
1) 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?
2) 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?
Thanks
To find the probabilities for the given scenarios, we can use the concept of probability and the properties of independent events.
1) Probability of completing at least five visits before needing a new supply:
To calculate this probability, we need to find the probability that Marie does not need a new supply on each visit, at least until the fifth visit.
Let's break down the events for each visit:
- The probability that the door is answered = 3/4
- The probability that there is a toddler in residence = 1/3
- The probability of neither event happening (door not answered or no toddler) = 1 - (3/4) * (1/3) = 7/12
Now, let's calculate the probability that Marie doesn't need a new supply for the first four visits:
P(Not needing a new supply on the 1st visit) = 1 (since she starts with 3 toys)
P(Not needing a new supply on the 2nd visit) = P(Not needing a new supply on the 1st visit) * P(Don't need a new supply on the 2nd visit)
= 1 * (7/12)
P(Not needing a new supply on the 3rd visit) = p(Not needing a new supply on the 2nd visit) * P(Don't need a new supply on the 3rd visit)
= (7/12) * (7/12)
P(Not needing a new supply on the 4th visit) = P(Not needing a new supply on the 3rd visit) * P(Don't need a new supply on the 4th visit)
= (7/12) * (7/12)
Now, let's calculate the probability that she completes at least five visits before needing a new supply:
P(Completing at least five visits before needing a new supply) = 1 - P(Not needing a new supply on the 1st, 2nd, 3rd and 4th visits)
= 1 - (7/12) * (7/12)
2) Expected value of the number of houses with toddlers visited without leaving any toys:
To find the expected value, we will calculate the average number of houses visited before needing a new supply, considering she starts with six toys. We need to find the probability of visiting houses without leaving any toys until she needs a new supply.
Let's break down the events for each visit:
- The probability that the door is answered = 3/4
- The probability that there is a toddler in residence = 1/3
- The probability of neither event happening (door not answered or no toddler) = 1 - (3/4) * (1/3) = 7/12
Now, let's calculate the probability that Marie visits a house without leaving any toys on a single visit:
P(Visit without leaving toys) = (3/4) * (2/3) = 1/2
(Choosing the toy-giving visit from the door answered and toddler in residence outcomes)
Since these events are independent between households, the probability of visiting a house without leaving any toys remains the same for each visit.
The expected value is given by:
Expected Value = Probability of visiting a house without leaving toys * Number of visits
Expected Value = (1/2) * Number of visits
Substituting the number of visits as the minimum required visits before needing a new supply (at least five visits), we get:
Expected Value = (1/2) * 5 = 2.5
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 2.5.