Suppose that the amount of time (in months) that a light bulb lasts before it burns out has an exponential distribution with parameter λ = 1/5. a) what is the probability that it lasts at least three months? b) if it has already lasted two months, what is the probability that it will last at least three more months? c) on average, how many months will the light bulb last?

We have

a)

Thus probability that the bulb lasts 3 months equals

b)

It is a case of conditional probability

Let A be the event that the bulb will last another 3 months provided it has lasted for 2 months

'B' be the event that that the light bulb lasts for 2 months

thus we have

also

=

c)

On an average the expectation value of the life of the bulb will be given by mean of the distribution

thus on an average the bulb will last for 5 months.

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.