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?
Let's calculate the probability that she does not distribute any toys on her first visit and on her second visit, then multiply these probabilities together to find the overall probability.
Probability of not distributing a toy on the first visit:
P(no toy distributed on 1st visit) = P(door not answered on 1st visit) + P(no toddler on 1st visit)
P(no toy distributed on 1st visit) = (1 - 3/4) + (1 - 1/3)
P(no toy distributed on 1st visit) = 1/4 + 2/3
P(no toy distributed on 1st visit) = 11/12
Now, let's calculate the probability that she does not distribute any toys on the second visit:
P(no toy distributed on 2nd visit) = P(door not answered on 2nd visit) + P(no toddler on 2nd visit)
P(no toy distributed on 2nd visit) = (1 - 3/4) + (1 - 1/3)
P(no toy distributed on 2nd visit) = 1/4 + 2/3
P(no toy distributed on 2nd visit) = 11/12
Now, let's find the overall probability that she has not distributed any toys by the end of her second visit:
P(no toys distributed by the end of 2nd visit) = P(no toy distributed on 1st visit) * P(no toy distributed on 2nd visit)
P(no toys distributed by the end of 2nd visit) = (11/12) * (11/12)
P(no toys distributed by the end of 2nd visit) = 121/144
P(no toys distributed by the end of 2nd visit) = 0.8403
Therefore, the probability that Marie has not distributed any toys by the end of her second visit is approximately 0.8403 or 84.03%.