Question

In seemmer, a truck driver get an average
Suncture in 1000 km . Find the probability he
will have (1)no peeneture (2)almost one
puncture in a journey of 3000km​

Answer 1

Answer:

I think so the answer is 2

Answer 2

Answer:1.e^-3

2.4e^-3

Explanation:

Let E be a event happening puncture of 1000kms...

Then...P(E)=1/1000

Then the mean punctures for 3000kms will be=np=3000*(1/1000)=3

Therefore ...mean,m=3

1.by Poisson distribution P(X=y)=

(m^y)(e^-m)/y!

Where...m is the mean

y is the number of tims the event

occured

Now.....for no punctures y=0

P(X=0)=(3^0)(e^-3)/0!=e^-3

Therefore the probability of no punctures is e^-3

2.probality of having atmost one puncture is........

P(X<=1)=P(X=0)+P(X=1)

=[(3^0)(e^-3)/0!]+[(3^1)(e^-3)/1!]

=(e^-3)+3e^-3

=4e^-3