Question

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 vis

2) What is the probability that she gives away the first toy on her fourth visit?

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?

4) What is the probability that she will give away the second toy on her fourth visit?

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?

6) 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?

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?

Answer 1

The probability of the door being answered is 3/4, and the probability that there is a toddler in residence is 1/3.

1) The probability of door opened and toddler on a visit =(3/4)*(1/3)=0.25

Therefore, the probability that she has not distributed any toys by the end of her second visit is:

(1-0.25)*(1-0.25) = 0.5625

2) The probability that she gives away the first toy on her fourth visit is:

(1-0.25)³*0.25 = 0.105469

3) Given that she has given away her second toy on her fifth visit, the conditional probability that she will give away her third toy on her eighth visit (means she did not give toys in sixth and seventh visits and give toy in eighth visit) is:

(1-0.25)*(1-0.25)*0.25 = 0.140625

4) The probability that she will give away the second toy on her fourth visit is:

³C₁(0.25)²(0.75)² = 0.105469