Question

If the number of orders at a production center this month is a Geom(0.7) random variable, find the probability that we'll have at most 3 orders.

Answer 1

Answer: 0.973

Step-by-step explanation:

The probability mass function of geometric distribution is

$f(x)=(1-p)^{x-1}p,\ \ \ x=1,2,3,...$, p= probability of success on a trial.

As per given, we have

p=0.7

The probability that we'll have at most 3 orders $P(X\leq3)=P(x=1)+P(x=2)+P(x=3)$

$=(1-0.7)^{1-1}(0.7)+(1-0.7)^{2-1}(0.7)+(1-0.7)^{3-1}(0.7)\\\\=0.7+0.21+0.063\\\\=0.973$

hence, the required probability = 0.973