Question

If 3% of the electric bulbs manufactured by a company are defective, find the probability that in a sample of 100 bulbs, (a) 0, (b) 1, (c) 2, (d) 3, (e) 4, (f) 5 bulbs will be defective.

Answer 1

a) 0.0476

b) 0.1471

c) 0.2252

d) 0.2275

e) 0.1706

f) 0.1013

Explanation:

For each bulb, there are only two possible outcomes. Either they are defective, or they are not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinatios of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem we have that:

There are 100 bulbs, so
`n = 100`.

3% are defective, so
`p = 0.03`.

a) 0

`P(X = 0) = C_(100,0).(0.03)^(0).(0.97)^(100) = 0.0476`

b) 1

`P(X = 1) = C_(100,1).(0.03)^(1).(0.97)^(99) = 0.1471`

c) 2

`P(X = 2) = C_(100,2).(0.03)^(2).(0.97)^(98) = 0.2252`

d) 3

`P(X = 3) = C_(100,3).(0.03)^(3).(0.97)^(97) = 0.2275`

e) 4

`P(X = 4) = C_(100,4).(0.03)^(4).(0.97)^(96) = 0.1706`

e) 5

`P(X = 5) = C_(100,5).(0.03)^(5).(0.97)^(95) = 0.1013`