Question

On each drive, you can have an accident with probability 0.2. Suppose each drive is independent of all previous ones. Let Y denote the number of accidents in six drives.

A) Find the probability distribution of Y. B) Expected value EY. C) Variance VY.

Answer 1

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(y) = ncy*(p^y)*(1-p)^n-y

Ncy = n!/(y!*(n-y)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.2

N = number of trials = n

Y = desired success

B)

Expected value is given by number of trials * P = n*0.2

C)

Varuance = n*p*(1-p) = n*0.2*0.8