Question

A certain office building infrequently experiences power outages. It is known that the probability of a power outage on any given day is 0.005, and power outages are independent of each other.
a) What is the probability that during any given 400 day period there will be one day with a power outage?
b) What is the probability that power outages will occur on more than three days during the 400 day period?

Answer 1

a) 400*.005
b) 400*.005*.005*.005 (Not sure on this one)

Answer 2

@whpalmer4

Answer 3

@perl

Answer 4

@whpalmer4

Answer 5

a) 400 *.005

Answer 6

b) Let's define a variable X = # of days of power outtage in 400 days.
so X can equal 0,1,2,3... 400
we want probability P ( X > 3 )

Answer 7

Now the complement is much easier since P(X>3) = P( X = 4) + P(X=5) + ... P(X = 400).
But
P(X>3) = 1- P(X<=3 )
= 1 - ( P(x=0) + P(x=1) +P(x=2) +P(x=3) )

Answer 8

I did part a like this == pr= (1/400)*0.005

Answer 9

it should be, n*p

Answer 10

400 * .005

Answer 11

@perl 1-(1/400+2/399+3/398)x0.005 for part B

Answer 12

this is a binomial probability problem
P(S) = .995
P(F) = .005
we want
1 - [ P( X = 0 ) + P( X = 1) + P( X = 2) + P( X = 3) ]
1 - ( .995^400 + (400 choose 1) * .995^399 * .005 + (400 Choose 2)*.995^398*.005^2+ (400 choose 3) * .995^397 * .005^3 )

Answer 13

@whpalmer4

Answer 14

What are you tagging me for when @perl is right there helping you?

Answer 15

it dose not make sense , cuz he is multiply success and failure prpbabilities with each other in same time ,

Answer 16

@whpalmer4

Answer 17

I have to leave, perhaps @mathmale can help you.

Answer 18

kalyan, this is a similar question http://users.rowan.edu/~schultzl/TI/binomial_StatI.pdf

Answer 19

sorry i dont understand your question

Answer 20

b) 1 - binomcdf( 400, .005, 3) = .1424