Question

On a given day, each computer in the 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. Find the probability that on any given day, there are;
(a) exactly 3 crashes
(b) at most 3 crashes
(c) at least 3 crashes
(d) more than 3 and less than 6.

Answer 1

X=number of computers crash in a day

Assume there are n number of computers

Distribution is,

$X \backsim binomial(n,0.05)$ .

a)

$P(X=3)=\binom{n}{3}*P^3*(1-P)^{(n-3)}\\ P(X=3)=\binom{n}{3}*0.05^3*0.95^{(n-3)}\\ \bold{P(X=3)=\frac{n(n-1)(n-2)}{120}*0.95^{(n-3)}}\\$

b)

$P(X\le 3)=\sum_{i=0}^{3}\binom{n}{i}*P^i*(1-P)^{(n-i)}\\ \bold{P(X\le 3)=\sum_{i=0}^{3}\binom{n}{i}*0.05^i*0.95^{(n-i)}}\\$

c)

$P(X\ge3)=1-P(X\le2)\\ P(X\ge 3)=1-\sum_{i=0}^{2}\binom{n}{i}*P^i*(1-P)^{(n-i)}\\ \bold{P(X\ge 3)=1-\sum_{i=0}^{2}\binom{n}{i}*0.05^i*0.95^{(n-i)}}\\$

d)

$P(3< X<6)=\sum_{i=4}^{5}P(X=i)\\ P(3< X< 6)=\sum_{i=4}^{5}\binom{n}{i}*P^i*(1-P)^{(n-i)}\\ P(3< X< 6)=\binom{n}{4}*0.05^4*0.95^{(n-4)}\\ \hspace{10 em}+\binom{n}{5}*0.05^5*0.95^{(n-5)}\\ \bold{P(3< X< 6)=\frac{n(n-1)(n-2)(n-3)}{480}*0.95^{(n-4)}}\\ \bold{\hspace{7 em}+\frac{n(n-1)(n-2)(n-3)(n-4)}{2400}*0.95^{(n-5)}}$

Here number of computers should be give.By substituting that value those probabilities can be found.