- What is the probability of that the first order will cause on 4th sales call of the day?

- If 8 sales calls are made on a day, what is the probability of getting exactly 6 orders?

- If 4 sales calls are made before lunch, what is the probability that one or less will result in an order.

• In first question we have to find the probability of that first order will cause on 4th sales call of the day.

  Here there is geometric distribution.

  p=48%=0.48

  q = 1-p = 1-0.48 = 0.52

  we have to determine first order will cause on 4th sales call hence x=4

  According to geometric distribution

  p(X=x) = p*q(x-1)

  p(X=4) = (0.48)*(0.52)(4-1)

  =0.067

• In the second question we have to find probability that exactly 6 order cause from 8 sales calls.

  Here x=6, n=8.

  As n<30 we use Binomial distribution.

  According to binomial distribution.

  p(X=x) =($^nC_x$) * ($p^x$) * ($q^{n-x}$)

  p(X=6) = ($^8C_6$) * ($0.48^6$) * ($0.52^{8-6}$)

  =0.092

• In last question we have to find the probability of that one or less result in order from 4.

  So n=4, x=1 or x=0

  p(X=0)+p(X=1)

  = ($^4C0$)($0.48^0$)($0.52^{(4-0)})+(^4C1$)($0.48^1$)($0.52^{(4-1)})$

  =0.343