- Suppose people arrive at a bank with poisson rate \(\lambda = 4\) per hour.
- What is the probability that 5 people arrive in the first half hour?
- What is the probability that at least 3 people arrive in the first hour?

**Solution**

- \(P(X = 5|\lambda t = 4*0.5)= \dfrac{e^{-\lambda t}(\lambda t)^5}{5!}\)

```
#pdf of poisson
dpois(5,lambda=4*0.5)
```

`## [1] 0.03608941`

- \(P(X \ge 3|\lambda t = 4) = 1 - P(X \le 3|\lambda t = 4) = 1 - \sum_{i=0}^3 \dfrac{e^{-\lambda t}(\lambda t)^i}{i!}\)$

```
#pdf of poisson
1 - ppois(3,lambda=4)
```

`## [1] 0.5665299`

Patients arrive at the doctorâ€™s office according to Poisson distribution with \(\lambda = 4\)/hour.

- What is the probability of getting less than or equal to 8 patients within 2 hours?
- Suppose each arriving patient has 25% chance to bring a person to accompany. There are 20 seats in the waiting room. At least many hours should pass that there is at least 50% probability that the waiting room is filled with patients and their relatives?

- \(P(X\le 8|\lambda t = 4*2)= \sum_{i=0}^8 \dfrac{e^{-\lambda t}(\lambda t)^i}{i!}\)

```
#cdf of poisson
ppois(8,lambda=4*2)
```

`## [1] 0.5925473`

- First let's define the problem. Define \(n_p\) as the number of patients and \(n_c\) is the number of company. We want \(n_p + n_c \ge 20\) with probability 50% or higher for a given \(t^*\). Or to paraphrase, we want \(n_p + n_c \le 19\) w.p. 50% or lower.

What is \(n_c\) affected by? \(n_p\). It is actually a binomial distribution problem. \(P(n_c = i|n_p) = \binom{n_p}{i} (0.5)^i*(0.5)^{n_p-i}\). It is even better if we use cdf \(P(n_c \le k|n_p) = \sum_{i=0}^{k} \binom{n_p}{i} (0.5)^i*(0.5)^{n_p-i}\).

We know the arrival of the patients is distributed with poisson. So, \(P(n_p = j|\lambda t^*) = \dfrac{e^{-\lambda t}(\lambda t)^j}{j!}\). So \(P(j + k \le N) = \sum_{a=0}^j P(n_p = a|\lambda t^*)*P(n_c \le N-a | n_p = a)\). Remember it is always \(n_c \le n_p\).

```
#Let's define a function
calculate_probability<-function(N=19,t_star=1,lambda=4,company_prob=0.25){
#N is the max desired number of patients
the_prob<-0
for(n_p in 0:N){
the_prob <- the_prob + dpois(n_p,lambda=lambda*t_star)*pbinom(q=min(N-n_p,n_p),size=n_p,prob=company_prob)
}
return(the_prob)
}
#Try different t_stars so probability is below 0.5
calculate_probability(t_star=3)
```

`## [1] 0.8380567`

` calculate_probability(t_star=3.3)`

`## [1] 0.7443518`

` calculate_probability(t_star=3.955)`

`## [1] 0.49914`

Suppose the the pdf of a random variable \(x\) is \(f(x) = \dfrac{a}{(1-x)^{1/3}}\) for \(0 < x < 1\) and \(0\) for other values of x. (Note: There was a typo in the in-class exercise saying the interval is \(0 < x < 1\))

- Find the constant \(a\).
- Find cdf of F(X < 3/4).

Integral of \(f(x)\) should be 1. So \(\int_0^1 f(x) dx = a*(-3/2(1-x)^{2/3})|_0^1 = 3a/2\). Then \(a = 2/3\).

Use the same integral \(\int_0^{3/4} f(x) dx = -(1-x)^2/3|_0^{3/4} = 1 - (1/4)^{2/3} = 0.603\)

Let \(X\) and \(Y\) be the random variables and \(f(x,y)\) is the probability density function of the joint distribution. Suppose \(f(x,y) = a(\dfrac{5x}{7} + \dfrac{9y^3}{2})\) if \(0<x<2\) and \(-1<y<1\) (0 otherwise).

- Find \(a\).
- Find the marginal distribution of \(y\) (\(h(y)\)) and \(h(y<0.5)\).
- Find the conditional distribution of \(f(y|x)\).

a) \(\int^1_{-1} \int^2_0 a(\dfrac{5x}{7} + \dfrac{9y^3}{2}) dx dy = \int^1_{-1} a(\dfrac{5x^2}{14} + \dfrac{9xy^3}{2})dy|^2_0 = \int^1_{-1} a(\dfrac{10}{7} + 9y^3) dy\). (This is also \(h(y)\) if \(a\) is known.) \(a(\dfrac{10y}{7} + \dfrac{9y^4}{4})|^1_{-1} = a(20/7)\). In order to be a distribution it should be equal to 1. So \(a = 7/20\).

b) As given in (a) \(h(y) = 7/20*(\dfrac{10}{7} + 9y^3)\). So \(h(y<0.5) = \int^{0.5}_{-1} 7/20*(\dfrac{10}{7} + 9y^3) dy = 7/20*(\dfrac{10y}{7} + \dfrac{9y^4}{4})|^{0.5}_{-1} = 0.0335\)

c) We need to find \(g(x) = \int^1_{-1} 7/20(\dfrac{5x}{7} + \dfrac{9y^3}{2})dy = x/2\). \(f(y|x) = f(x,y)/g(x)\).