Question

Screws are sold in packets of 15. Faulty screws occur randomly. A large number of packets are tested 

for faulty screws and the mean number of faulty screws per packet is found to be 1.2.

(i) Show that the variance of the number of faulty screws in a packet is 1.104.

(ii) Find the probability that a packet contains at most 2 faulty screws.

Damien buys 8 packets of screws at random.

(iii) Find the probability that there are exactly 7 packets in which there is at least 1 faulty screw.

Answer 1

Binomial distribution

i)

$E(X)=np=15p=1.2$

$p=1.2/15=0.08$

$Var(X)=np(1-p)=1.2\cdot0.92=1.104$

ii)

$P(x\le2)=P(0)+P(1)+P(2)=$

$=0.92^{15}+C^{1}_{15}\cdot0.08\cdot0.92^{14}+C^{2}_{15}\cdot0.08^2\cdot0.92^{13}=$

$=0.2863+0.3734+0.2273=0.887$

iii)

The probability that a packet contains at least 1 faulty screw:

$P(x\ge1)=1-P(0)-P(1)=1-0.2863-0.3734=0.3403$

The probability that there are exactly 7 packets in which there is at least 1 faulty screw:

$P(y=7)=C^7_8\cdot0.3403^7\cdot(1-.03403)=0.0028$