Question

You manage a pizzeria. According to an independent study, your deliveries have been late an average of 5%. What is the probability that the 9th delivery of the day will be late? What Probability Distribution did you use?

Answer 1

Answer:

We will use geometric distribution.

The required probability is 0.0331

Step-by-step explanation:

Consider the provided information.

Deliveries have been late an average of 5%.

The geometric distribution gives the probability that k independent trials are required for the first occurrence of success, each with the probability of success p. If the probability of success in each trial is p, then the probability of the first success being the kth trial (out of k trials).

${\displaystyle \Pr(X=k)=(1-p)^{k-1}p}$

The probability that the 9th delivery of the day will be late is:

required probability = $0.05\times(1 - 0.05)^8 = 0.0331710216$

Hence, the required probability is 0.0331