Question 566685: From experience, Teton Tire knows the probability is .95 that a particular XB-70 tire will last 60,000 miles before it becomes bald or fails. An adjustment is made on any tire that does not last 60,000 miles. You purchase 4 XB-70’s.
a. (10 pts) What is the probability that all 4 tires will last at least 60,000 miles?
b. (10 pts) What is the probability that none of the tires will last at least 60,000 miles?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! probability that all 4 tires will last at least 60,000 miles is .95^4 = .81450625
probability that all 4 tires will not last at least 60,000 miles is 1 - .81450625 = .18549375
probability that none of the tires will last at least 60,000 miles is .05^4 = 6.25 * 10^-6
the total probabilities are shown below:
p(4 last 60,000) = 4C0 * .95^4 * .05^0 = 1 * .95^4 * .05^0 = .81450625
p(3 last 60,000) = 4C1 * .95^3 * .05^1 = 4 * .95^3 * .05^1 = .l71475
p(2 last 60,000) = 4C2 * .95^2 * .05^2 = 6 * .95^2 * .05^2 = .0135375
P(1 last 60,000) = 4C3 * .95^1 * .05^3 = 4 * .95^1 * .05^3 = 4.75 * 10^-4
p(0 last 60,000) - 4C4 * .95^0 * .05^4 = 1 * .95^0 * .05^4 = 6.25 * 10^-6
add all these probabilities together and you get a total probability of 1.
the probability that 4 last 60,000 is.81450625.
the probability that 0 last 60,000 is 6.25 * 10^-6.
the probability that 4 do not last 60,000 is equal to 1 - .81450625 = .18549375.
this is the probability that 0 and 1 and 2 and 3 last 60,000 which is the total probability minus the probability that 4 last 60,000 miles.
you were asked:
a. (10 pts) What is the probability that all 4 tires will last at least 60,000 miles?
answer is.81450625
b. (10 pts) What is the probability that none of the tires will last at least 60,000 miles?
answer is 6.25 * 10^-6 which is equivalent to .00000625
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