Answer to Question #106356 in Statistics and Probability for FFG

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

a. Let $<\Omega,F,P>$ is a probability space. Random events $A$ and $B$ from $F$ are said to be independent if $P(A\bigcap B)=P(A)\cdot P(B).$

b. i) Let random event $A$ is "the first customer is female", random event $B$ is "the second customer is female", random event $C$ is "the third customer is female". Then $P(A\bigcap B\bigcap C)=P(A)\cdot P(B)\cdot P(C)=(0.65)^3\approx 0.275$

(events $A, B \text{ and } C \text{ are collectively independent)}.$

ii) Probability that the customer is male equals $1-0.65=0.35.$

Let random event $A$ is "the first customer is male", random event $B$ is "the second customer is male", random event $C$ is "the third customer is male". Then $P(A\bigcap B\bigcap C)=P(A)\cdot P(B)\cdot P(C)=(0.35)^3\approx 0.043$

(events $A$, $B$ and $C$ are collectively independent).

iii) We have to find probability that at least one (the first, the second or the the third) customer is male:

$P=1-0.275=0.725$ (events "all three customers are female" and "at least one customer is male" are opposite).

c. Random events:

$A$ --- "did not have an accident"

$B$ --- "had an accident"

$C$ --- "had instructions"

$D$ --- "did not have instructions"

We have:

$P(A)=0.99\\ P(B)=0.01\\ P(C)=0.1\\ P(D)=0.9\\ P(C|B)=0.4\\ P(D|B)=0.6$

We should find $P(A|D)$ and $P(A|C).$

$P(A|D)=\frac{P(A)\cdot P(D|A)}{P(D)}\\ P(D)=P(A)P(D|A)+P(B)P(D|B)\\ 0.9=0.99P(D|A)+(0.01)(0.6)\\ P(D|A)\approx 0.903.\\ i) P(A|D)=\frac{(0.99)(0.903)}{0.9}=0.9933.\\ P(A|C)=\frac{P(A)\cdot P(C|A)}{P(C)}\\ P(C|A)=1-P(D|A)=1-0.903=0.097\\ ii) P(A|C)=\frac{(0.99)(0.097)}{0.1}=0.9603.$