The probability that a bulb will fuse within an year is 0.05. Find the
probability that out of 5 bulbs 3 bulbs will fuse within the year.
Answers
The probability that a bulb will fuse within an year is 0.05, then the probability that out of 5 bulbs 3 bulbs will fuse within the year is 10 x (1/20)^3 x (19/20)^2 = 0.001128125
Given:
The probability that a bulb will fuse within a year is 0.05
To find:
The probability that out of 5 bulbs 3 bulbs will fuse within the year
Solution:
Let p denote the probability that a bulb will fuse within a year and q denote the probability that a bulb will not fuse within a year
p = 0.05
q = 1 - p
=> q = 1 - 0.05
=> q = 0.95
Using the concept of binomial distribution,
p = probability of success = 0.05
q = probability of failure = 0.95
Therefore,
the probability that out of 5 bulbs 3 bulbs will fuse within the year
= 5C3 x (0.05)(0.05)(0.05) x (0.95)(0.95)
= 5C3 x (0.05)^3 x (0.95)^2
= 10 x (0.05)^3 x (0.95)^2
= 10 x (1/20)^3 x (19/20)^2
= 0.001128125
Hence,
The probability that out of 5 bulbs 3 bulbs will fuse within the year is 10 x (1/20)^3 x (19/20)^2 = 0.001128125
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