Question

A room has three lightbulbs. Each one has a 10% probability of burning out within the month. Write the probability as a percentage rounded to one decimal place.

What is the probability that all three will burn out within the month?

"If something can go wrong, it will go wrong." This funny saying is called Murphy's law. Let's interpret this to mean "If something can go wrong, there is a very high probability that it will eventually go wrong."

Suppose we look at the event of having an automobile accident at some time during a day's commute. Let's assume that the probability of having an accident on a given day is 1 in a thousand or 0.001. That is, in your town, one of every thousand cars on a given day is involved in an accident (including little fender-benders). We also assume that having (or not having) an accident on a given day is independent of having (or not having) an accident on any other given day. Suppose you commute 42 weeks per year, 5 days a week, for a total of 210 days each year. In the following parts, write each probability in decimal form rounded to three places.(a) What is the probability that you have no accident over a year's time?


(b) What is the probability that you have at least one accident over a one-year period?


(c) Repeat part (a) for a 10-year period and for a 20-year period.

10-year period
20-year period

Repeat part (b) for a 10-year period and for a 20-year period.

10-year period
20-year period

ALSO,

A room has three lightbulbs. Each one has a 10% probability of burning out within the month. Write the probability as a percentage rounded to one decimal place.

What is the probability that all three will burn out within the month?

Answer 1

Probability of burning out within a month is 0.1

Probability of all 3 burning out is product of the probability of burning out of all bulbs = 0.1*0.1*0.1=0.001

Proabability of having an accident on a paticulr day = 0.001
Therefore probability of not having an accident on a paticular day = 1 - 0.001 = 0.999

a) Probability of having no accident in the whole year (210 days) = product of probability of not having accident on a day. = (0.999)^210

b) Probability of having at least one accident in a year = 1 - probability of no accident in a year = 1 - (0.999)^210

c)for 10 year period, days = 210 *10 = 2100 so probability = (0.999)^2100
for 20 year period, days = 210 *20 = 4200 so probability = (0.999)^4200

d)for 10 year period, days = 210 *10 = 2100 so probability = 1 - (0.999)^2100
for 20 year period, days = 210 *20 = 4200 so probability = 1 - (0.999)^4200