How do I solve the questions below using the Poisson distribution with the variable $t$?

Question

This is the problem I have to solve for a job at school. Can anyone help me, what kind of distributions approximations do I use and :

In (a), how do I manipulate t in order to find the desired probability? The dependence of the intervals of emission of particles, what kind of information does it give us? How to handle t in seconds with parameter 228t / min?

In (b), what information gives us the fact that at the last minute no particles were emitted? and how to calculate the requested probability?

In (c), I do not know how to find the random variable for time in seconds from the poisson in order to address the time between two emissions!

The question

The number of particles emitted by a radioactive source over a range of time period I of duration t follows a Poisson distribution of 228t / min parameter, and is independent of the number of particles emitted during any other interval apart from I.

(a) Determine the probability that more than 4 particles are emitted in one second.

(b) Assuming that in the last minute no particles were emitted, determine the probability that within at least 5 seconds at least 2 particles are emitted.~

(c) Let T be the random variable representative of the time in seconds between two emissions. Determine P (T <2).

Answer 1

Here is what I hope is a useful outline, with some numerical answers from software for you to verify using formulas from a probability text.

(a) Let $X \sim \mathsf{Pois}(\lambda = 3.8)$ be the number of particles in one second. You seek $$P(X > 4) = 1 - P(X \le 3) \\= 1 -P(X=0)+P(X=1)+P(X=2)+P(X=3)\\ = 0.5265,$$ which can be evaluated using the Poisson PDF formula or by software. You should use the Poisson PDF to verify the answer. In R statistical software, where ppois is a Poisson CDF, we have

1 - ppois(3, 3.8)
##   0.5265152

(b) Because counts in disjoint time periods are independent, the past history does not matter. I find this question to be awkwardly worded. Here is my opinion of its interpretation. Let $Y \sim \mathsf{Pois}(\lambda = 3.8(5)),$ and you seek $P(Y > 2),$ which may be evaluated similarly to (a). The probability should be large because, on average, we expect 19 particles in 5 sec.

1 - ppois(1, 5*3.8)
## 0.9999999

(c) We begin by discussing the relationship between Poisson and exponential distributions.

Let $X_t \sim \mathsf{Pois}(\lambda t)$ be the number of particles seen in $t$ seconds. Then the probability of seeing no particles in time interval $(0, t]$ is $P(X = 0) = e^{-\lambda t}.$

Starting at time $0,$ let $T$ be the waiting time until the first particle arrives. Then $P(T > t)$ is another way to ask for the probability of seeing no particles in $(0, t].$ Thus, the CDF of $T$ is $$F_T(t) = P(T \le t) = 1 - P(T > t) = 1 - e^{-\lambda t},$$ for $t > 0.$ Taking the derivative, we have $$f_T(t) = F_T^\prime(t) =\lambda e^{-\lambda t},$$ which is the density function of the exponential distribution with rate $\lambda,$ denoted $T \sim \mathsf{Exp}(\lambda).$ One can show that $E(T) = 1/\lambda.$ If $\lambda = 3.8,$ then the average time between particle emissions is $1/3.8 = 0.2632$ sec. On average, we have to wait a little over a quarter of a second between particle emissions. You should fill in the details using calculus.

Now we seek $P(T > 2) = 1 - F_T(2) = 0.0005.$ You should verify this. Intuitively, if there are almost four particles a second, then the probability of having to wait more than two seconds for the first particle should be very small.

In R, where pexp is an exponential CDF, we have:

1 - pexp(2, 3.9)
##  0.0005004514

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Answer 2

This is my interpretation of the questions

a) they mean $228$ per minute so $3.8$ per second. Therefore calculate $$p(X>4)=1-p(X\leq 4)$$ using Poisson with $\lambda=3.8$

b) assuming they mean at least $2$ particles in $5$ seconds (since each period is time is independent, you may ignore the condition about no particles in the last minute) now use Poisson with $\lambda=5\times3.8$. You therefore have the option of a Normal approximation.

c) the waiting time between Poisson events follows an exponential distribution. You might like to check this out for yourself.

I hope this helps.