Help s2 urgent

2. A company receives telephone calls at random at a mean rate of 2.5 per hour.
(a) Find the probability that the company receives
(i) at least 4 telephone calls in the next hour,
(ii) exactly 3 telephone calls in the next 15 minutes.
(5)
(b) Find, to the nearest minute, the maximum length of time the telephone can be left
unattended so that the probability of missing a telephone call is less than 0.2
(3)
The company puts an advert in the local newspaper. The number of telephone calls
received in a randomly selected 2 hour period after the paper is published is 10
(c) Test at the 5% level of significance whether or not the mean rate of telephone calls
has increased. State your hypotheses clearly.
(5)

I don't get part b,
I solved it this way

P(X=0)<0.2

BUT the mark scheme says that it should be 1-P(X=0)<0.2
why did the subtract from 1?

A call will be missed if X > 0.

So you want P(X>0) < 0.2

And P(X>0) = 1-P(X=0)

Hence the requirement 1-P(X=0) < 0.2.

X is the number of telephone calls received right?

A call will be missed if X > 0.

So you want P(X>0) < 0.2

And P(X>0) = 1-P(X=0)

Hence the requirement 1-P(X=0) < 0.2.

I'm sorry I still don't get it

OK, The random variable X represents the number of telephone calls received in a given (unknown) time period.

You will miss a call if X>0 in that time period - the phone not being attended.

You want the probability of missing a call to be less than 0.2, so you want P(X>0)<0.2

but doesn't P(X>0) means that the probability of receiving more than 0 calls?

Yes.

If you receive more than 0 calls, you will miss a call.

okay I got that bit
but then essentially we are saying that P(X>0)<0.2 means that the probability of telephone calls received should be less than 0.2, how does that connect to missing a phone call?

I'm sorry for bothering you with this and I appreciate all the help you've given me! thanks again

We're working out how long we can leave the phone unattended.

The r.v. X relates to the number of phone calls received during that time when the phone is unattended. We don't know how long that is, and that's what we're trying to work out.

Hence the P(X>0)<0.2

Rearranging it, as before, and using the definition of the poisson probability we can determine the parameter that makes it true. That parameter is proportional to the time we can leave the phone unattended.