Yes it’s indeed a binomial distribution

Here is how it would go : You can receive 180 calls evenly distributed across the time line

Now probability of receiving a call , p = 1/3. Not receiving a call , 1-p = 2/3

Total independent events = 180( Total number of calls ). n = 180.

k represents the number of calls received , so probability that 50 calls were received

= (180 Choose 50)*§^50*(1-p)^130

Now you need probability from 50 to 70 so you would need to sum the above probability for 21 cases that is when k goes from 50 to 70.

It can be done is excel by generating numbers from 50 to 70. In next column calculate the probability for each K.

The sum the values of the probability column.

Here is some R code to do the above:

```
num <- seq(50,70,1)
num
[1] 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
prob <- choose(180,num)*(1/3)^(num)*(2/3)^(180-num)
head(prob)
[1] 0.01825059 0.02326056 0.02885204 0.03484020 0.04096949 0.04692869
sum(prob)
[1] 0.9035098
```

So answer is 0.9035098