Question

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Stat

(1)

An insurance company believes that people can be divided into two classes: those who are an accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at some time within a fixed 1-year period with probability .4, whereas this probability decreases to .2 for a person who is not accident-prone.
(i) If we assume that 30 percent of the population is accident prone, what is the probability that a new policyholder will have an accident within a year of purchasing a policy?
(ii) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident-prone?

(2)

Two boxes containing marbles are placed on a table. The boxes are labeled B1 and B2. Box B1 contains 7 green marbles and 4 white marbles. Box B2 contains 3 green marbles and 10 yellow marbles. The boxes are arranged so that the probability of selecting box B1 is 1/3 and the probability of selecting box B2 is 2/3 FATIMAH is blindfolded and asked to select 3 marbles. She will win a color TV if she selects a green marble.
(i) What is the probability that FATIMAH will win the TV (that is, she will select a green marble)?
(ii) If FATIMAH wins the color TV, what is the probability that the green marble was selected from the first box?

(3)

One-half percent of the population has CORONA Virus. There is a test to detect CORONA. A positive test result is supposed to mean that you have CORONA but the test is not perfect. For people with CORONA, the test misses the diagnosis 2% of the times. And for the people without CORONA, the test incorrectly tells 3% of them that they have CORONA.
(i) What is the probability that a person picked at random will test positive?
(ii) What is the probability that you have CORONA given that your test comes back positive?

(4)

A device is composed of two components, A and B, subject to random failures. The components are connected in parallel and, consequently, the device is down only if both components are down. The two components are not independent. We estimate that the probability of:
a failure of component A is equal to 0.2;
a failure of component B is equal to 0.8 if component A is down;
a failure of component B is equal to 0.4 if component A is active.
(a)

Calculate the probability of a failure
(i) of component A if component B is down
(ii) of exactly one component
(b)

In order to increase the reliability of the device, a third component, C, is added in such a way that components A, B, and C are connected in parallel. The probability that component C breaks down is equal to 0.2, independently of the state (up or down) of components A and B. Given that the device is active, what is the
probability that component C is down?

Answer 1

Solution :

1)

i) Let,

Probability of finding a person who is accident prone P(A) = 0.30

Probability of finding a person who is not accident prone P(B) = 1 - 0.30 = 0.70

Probability of given person is accident prone and that he will face accident in certain period P(X) = 0.40

Probability of given person is not accident prone and that he will face accident in certain period P(Y) = 0.20

Probability that a new policyholder will have an accident within a year of purchasing a policy = P(Z)

Therefore,

P(Z) = P(A)*P(X) + P(B)*P(Y)

= 0.30*0.40 + 0.70*0.20

P(Z) = 0.26 = 26%

ii) Probability that policy holder is accident-prone = P(M)

Therefore,

P(M) = (P(A)*P(X))/ P(Z)

= (0.30*0.40)/ 0.26

P(M) = 0.4615 = 46.15%

2)

i) Given :

Box B1 contains 7 green and 4 white marbles = 11 marbles

Box B2 contains 3 green and 10 yellow marbles = 13 marbles

Probability of selecting Box B1 is 1/3 i.e. P(B1) = 1/3

Probability of selecting Box B2 is 1/3 i.e. P( B2) = 2/3

Let,

the probability that FATIMAH will win the TV (that is, she will select a green marble) = P(A)

Therefore,

P(A) = P(A/B1)*P(B1) + P(A/B2)*P(B2)

= (7/11)*(1/3) + (3/13)*(2/3)

P(A) = 0.3660 = 36.60%

ii)

Box B1 contains 7 green and 4 white marbles = 11 marbles

Probability of green marble drawn for Box B1 = P(G) = 7/11

Probability of selecting Box B1 is 1/3 i.e. P(B1) = 1/3

Let,

Probability of Box B1 and Green marble = P(X)

Therefore,

P(X) = P(G)*P(B1)

= (7/11) * (1/3)

P(X) = 0.2120 = 21.20%