Question

If the probability that a fluorescent light has a useful life of at least 500 hours is 0.8, find the probability that among 20 such lights

a) exactly 18 will have a useful life of at least 500 hours
b) at least 15 will have a useful life of at least 500 hours

Answer 1

a)Probability( exactly 18 will have a useful life of at least 500 hours

)= 0.1368

b) Probability(at least 15 will have a useful life of at least 500 hours)=0.8042

Explanation:

Let X be random variable.It represent the number of light having a usefull life of at least 500 hours 20 fluorescent lights

Lets consider it success if a light has a useful life of at least 500 hours . so

Probability of a success in each trial is p =0.8

Because of the trials are independent, X has Binomial distribution with parameters n= 20 and p=0.8

The probability mass function is,

P(X=x)=b(x;20,0.8)

where x=0,1,2,3,4....20

=
`\left(\begin{array}{c}{20} \\ {x}\end{array}\right)(0.08)^x(1-0.80)^(20-x)`

`=\left(\begin{array}{c}{20} \\ {x}\end{array}\right)(0.08)^x(0.20)^((20-x))`................(1)

A)exactly 18 will have a useful life of at least 500 hours

We Have to find the probability that exactly 18 will have a useful life of at least 500 hours.

using equation (1)

P(X= 18) =b(18,20,0.08)

P(X= 18)=
`\left(\begin{array}{c}{20} \\ {x}\end{array}\right)(0.80)^18(0.20)^((20-18))`

P(X= 18)=
`(20!)/(18!2!)(0.80)^(18)(0.20)^2`

P(X= 18)=
`190* 0.0180 * 0.04`

P(X= 18)=0.1368

B) at least 15 will have a useful life of at least 500 hours

We Have to find the probability that at least 15 will have a useful life of at least 500 hours

P(X>= 15) = 1-P(X<=14)

P(X>= 15)=1-
`\sum_(x=0)^(4) b(x ; 20,0.80)`

P(X>= 15) =
`1-[(20!)/(14!6!)(0.80)^(14)(0.20)^6+(20!)/(13!7!)(0.80)^(13)(0.20)^7+............+(20!)/(1!19!)(0.80)^(1)(0.20)^(19) +(20!)/(0!20!)(0.80)^(0)(0.20)^(20)]`

P(X>= 15)=
`1-[0.109+0.0545+0.0222+0.0074+0.0020+0.0005+0.0001+0+0+0+0+0+0+0+0]`

P(X>= 15)1= -0.1958

P(X>= 15)=0.8042