Question

Q1. On a given day, each computer in a lab has at most one crash. There is a 5% chance that a
computer has a crash during the day independent of the performance of any other computers in the
lab. There are 25 computers in the lab. Find the probability that on a given day, there are;
i Exactly 3 crashes.
ii At most 3 crashes

Answer 1

We have that

P(crash) = 0.05

n = 25

This follows binomial distribution.

The binomial probability is calculated by the formula:

$$P(X=m)=C(n,m)\cdot p^m \cdot (1-p)^{n-m}$$

i. m = 3

$P(X=3)=C(25,3)\cdot 0.05^3 \cdot (1-0.05)^{25-3}=\frac{25!}{3!22!}\cdot0.05^3\cdot0.95^{22}=0.093$

ii. m ≤ 3

$P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X = 3)$

$P(X=0)=C(25,0)\cdot 0.05^0 \cdot 0.95^{25}=0.277$

$P(X=1)=C(25,1)\cdot 0.05^1 \cdot 0.95^{24}=0.365$

$P(X=2)=C(25,2)\cdot 0.05^2 \cdot 0.95^{23}=0.231$

$P(X\le3)=0.277+0.365+0.231+0.093=0.966$

Answer:

I. 0.093

ii. 0.966