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)=[tex](0.995)^{100}[/tex]=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%.
A) In a string of 100 lights all 100 will last 2 years is 0.61.
B) Probability that at least one bulb will burn out in 2 years is 0.39.
Probability is the quantification of possibilities or chances.
It is given that,
The probability a single bulb will last 2 years is 0.995.
So, the probability that in a string of 100 lights all 100 will last 2 years will be = 0.995^100 = 0.61
So, the probability of at least one bulb will burn out will be equivalent to the probability that all 100 bulbs will burn out in 2 years i.e. 1-0.61 = 0.39
Therefore, A) In a string of 100 lights all 100 will last 2 years is 0.61.
B) Probability that at least one bulb will burn out in 2 years is 0.39.
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