A system is composed of 5 identical independent elements in parallel. What should be the reliability of each element to achieve a
reliability of 0.96 ?
Answer:1-((1-0.96))^(1/5))=1-0,52=0,48
A has one share in a lottery in which there is one prize and two blanks ; B has three shares in a lottery in which there are three prizes and 6 blanks; compare the probability of A's success to that of B's success.
Answer:
xCy= x!/(y!(x-y)!)
Probability A to loss =6C3/9C3
Probability A to win = 1−9C3/6C3=1−6!/(3!*3!)×(3!*6!)/9!=1−120/504=384/504=16/21
Probability B to loss = 2C1/3C1
Probability B to win = 1−(2C1/3C1)=1−2/3=1/3
required ratio is
16/21 : 1/3=16 : 7
A man and his wife appear in an interview for two vacancies in the same post.
The probability of husband's selection is 1/7
and the probability of wife's selection is 1/5
What is the probability that only one of them is selected ?
Answer:
P(A)=1/7
P(B)=1/5
So there is 6/7 probability for men to not to be selected and 4/5 for woman
So required probability will be:
P[(A and notB)or(B and notA)]=(1/7)*(4/5)+(1/5)*(6/7)=(4+6)/35=10/35
Four persons are chosen at random from a group containing 3 men, 2 women and 4 children.
Calculate the chances that exactly two of them will be children.
Answer:
xCy= x!/(y!(x-y)!)
All ways of groups:
9C4=126
Because we want 2 of choosen to be a children:
4C2=6
and choose the other 2 people will give:
5C2=10
So the probability will be (10*6)/126 = 10/21
A cinema house gets electric power from a generator run by diesel engine. On any day,
the probability that the generator is down (event A) is 0.025 and the probability
that the diesel engine is down (event B) is 0.04. What is the probability that the
cinema house will have power on any given day? Assume that occurrence of
event A and event B are independent of each other.
Answer:
At first we will calculate the probabilyty of losing power in cinema
P(AUB)=P(A)+P(B)-P(A)*P(B)=0.025+0.04-0.025*0.04=0.064
so the answer will be 1-0.064=0.936