Chapter 9 Probability Ex 9.4
Chapter 9 Probability Ex 9.4
Question 1.
There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen at random and marble is picked from the chosen bag. What is the probability that the chosen marble is red?
Solution:
Let event R: Chosen marble is red.
Let event Bi: ith bag is chosen.
∴ P(Bi) = 1/3
If Bag 1 is chosen, it has 75 red and 25 blue marbles.
Question 2.
A box contains 2 blue and 3 pink balls and another box contains 4 blue and 5 pink balls. One ball is drawn at random from one of the two boxes and it is found to be pink. Find the probability that it was drawn from
(i) first box
(ii) second box
Solution:
(i) By Bayes’ theorem,
the probability that a pink ball is drawn from the first box, is given by
(ii) By Bayes’ theorem,
the probability that a pink ball is drawn from the second box, is given by
Question 3.
There is a working women’s hostel in a town, where 75% are from neighbouring town. The rest all are from the same town. 48% of women who hail from the same town are graduates and 83% of the women who have come from the neighbouring town are also graduates. Find the probability that a woman selected at random is a graduate from the same town.
Solution:
Let the total number of women be 100.
∴ n(S) = 100
Let event N: Women are from neighbouring town,
event W: Women are from same town and
event G: Women are graduates.
Number of women from neighbouring town,
n(N) = 75
Number of women from same town,
By Bayes’ theorem, the probability that a women selected at random is a graduate from the same town, is given by
Question 4.
Question 5.
Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn from Jar I, and if it is Tail, a ball is drawn from Jar II. Suppose that this experiment is done and a white ball was drawn. What is the probability that this ball was in fact taken from Jar II?
Solution:
Question 6.
A diagnostic test has a probability 0.95 of giving a positive result when applied to a person suffering from a certain disease, and a probability 0.10 of giving a (false) positive result when applied to a non-sufferer. It is estimated that 0.5% of the population are sufferers. Suppose that the test is now administered to a person about whom we have no relevant information relating to the disease (apart from the fact that he/she comes from this population). Calculate the probability that:
(i) given a positive result, the person is a sufferer.
(ii) given a negative result, the person is a non-sufferer.
Solution:
Let event T: Test positive
event S: Sufferer
∴ P(S’) = 1 – P(S) = 1 – 0.005 = 0.995
Since a probability of getting a positive result when applied to a person suffering from a disease is 0.95 and probability of getting positive result when applied to a non sufferer is 0.10.
∴ P(T/S) = 0.95 and P(T/S’) = 0.10
∴ P(T) = P(S) P(T/S) + P(S’) P(T/S’)
= 0.005 × 0.95 + 0.995 × 0.10
= 0.10425
∴ P(T’) = 1 – P(T) = 1 – 0.10425 = 0.8958
(i) Required probability = P(S/T)
By Bayes’ theorem,
(ii) P(T’/S’) = 1 – 0.1 = 0.9
Required probability = P(S’/T’)
By Bayes’ theorem
Question 7.
A doctor is called to see a sick child. The doctor has prior information that 80% of the sick children in that area have the flu, while the other 20% are sick with measles. Assume that there is no other disease in that area. A well-known symptom of measles is rash. From the past records, it is known that, chances of having rashes given that sick child is suffering from measles is 0.95. However occasionally children with flu also develop rash, whose chance are 0.08. Upon examining the child, the doctor finds a rash. What is the probability that child is suffering from measles?
Solution:
Let the total number of sick children be 100.
∴ n(S) = 100.
Let event A: The child is sick with flu,
event B: The child is sick with measles,
event C: The child is sick with rash.
Required probability = P(B/C)
By Bayes’ theorem,
Question 8.
Solution:
= 0.02 × 0.9 + 0.1 × 0.6 + 0.88 × 0.1
= 0.166
Required probability = P(A1/B)
By Baye’s theorem
Question 9.
A box contains three coins: two fair coins and one fake two-headed coin. A coin is picked randomly from the box and tossed.
(i) What is the probability that it lands head up?
(ii) If happens to be head, what is the probability that it is the two-headed coin?
Solution:
Let event A: Fair coin is tossed,
event B: Fake coin is tossed
and event H: Head occur.
Clearly, a fair coin has one head.
∴ Probability that head occur under the condition that the fair coin is tossed = P(H/A) = 1/2
Fake coin has two heads.
∴ Probability that head occur under the condition that the fake coin is tossed = P(H/B) = 1
(ii) Required probability = P(B/H)
By Baye’s theorem
Question 10.
Required probability = P(C/E)
By Baye’s theorem
Question 11.
Solution:
Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively.
Let L be event that he is late.
