Answer to Question #106101 in Statistics and Probability for Eric Eshun

Question

State clearly by what is meant by two events being statistically independent.
b. In 2015, a company estimated that 65% of its customers were female and, in that year, three customers
entered one of their shops. Assuming that these customers were not related or acquaintances, find the
probability that they were:
i. all female
ii. all male
iii. at least one male
c. In a certain factory which employs 10,000 men, 1% of all employees have a minor accident in a given
year. Of these, 40% had safety instructions whereas 90% of all employees had no instructions. What is the
probability of an employee being accident free:
i. Given that he had no safety instructions
ii. Given that he had safety instructions

Answer 1

Linearly independent means one event has not any effect on the other event. In another way there is not any interception of the two event. In mathematically if A and B are linearly independent then,

$$P(A \cap B)=P(A).P(B)\\$$

b. Let event A=person is female & since entrance persons are not linearly dependent

i. $P(A)*P(A)*P(A)=0.65*0.65*0.65=0.2746$

ii $(1-P(A))*(1-P(A))*(1-P(A))=0.35*0.35*0.35=0.0429$

iii $P(at least one male)=1-P(all female)=1-0.2746=0.7254$

c.Let events,

A=has accident

B=has no accident

M=employees had instructions

N=employees hadn't instructions

$P(A)=0.01\\ P(B)=0.99\\ P(M|A)=0.4\\ P(N)=0.9\\ P(M)=0.1$

let ,

$P(N|A)=1-P(M|A)=0.6\\$

$P(N|B)=p;\\ P(N)=P(N|A).P(A)+P(N|B).P(B)\\ 0.9 =0.6*0.01+p*0.99\\ p=0.903$

i. An employee being accident free given that he had no safety instructions $P(B|N),$

$P(B|N)=\frac{P(N|B).P(B)}{P(N)}\\ P(B|N)=\frac{0.903*0.99}{0.9}\\ P(B|N)=0.9933$

ii. An employee being accident free given that he had safety instructions P(B∣M),

$since, P(M|B)=1-P(N|B)\\ P(M|B)=1-0.903=0.097\\ P(B|M)=\frac{P(M|B).P(B)}{P(M)}\\ P(B|M)=\frac{0.097*0.99}{0.1}\\ P(B|M)=0.9603$