Question

Christmas lights are often designed with a series circuit. This means that when one light burns out, the entire string of lights goes black. Suppose the lights are designed so that the probability a bulb will last 2 years is 0.995. The success or failure of a bulb is independent of the success or failure of the other bulbs.A) What is the probability that in a string of 100 lights all 100 will last 2 years?B) What is the probability at least one bulb will burn out in 2 years?

Answer 1

Answer:

a) The probability that in a string of 100 lights all 100 will last 2 years is 0.6058 (60.58%).

b) The probability at least one bulb will burn out in 2 years is 0.3942 (39.423%)

Step-by-step explanation:

a) We know that the probability of one bulb will last 2 years is 0.995. As the success or failure of a bulb is independent of the success or failure of the other bulbs, the probability of 100 bulbs is equal to the multiplication of each probability of each bulb (Multiplication Rule).

P(100 bulbs will last 2 years)=$(0.995)^{100}$=0.6058

Hence, the probability that in a string of 100 lights all 100 will last 2 years is 0.6058 or 60.58%.

b) When we say the probability at least one bulb will burn out in 2 years, is equal to the contrary event of 100 bulbs will last 2 years. (one or more bulbs will burn out)

P(at least one bulb will burn out in 2 years)= 1-P(100 bulbs will last 2 years)

P(at least one bulb will burn out in 2 years)= 1-0.6058= 0.3942

Hence, the probability at least one bulb will burn out in 2 years is 0.3942 or 39.423%.