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Solution:
Let us assume that E be the event of getting a defective item.
So, we have
P(A) = 50 % = 1/2
P(B) = 30 % = 3/10
P(C) = 20 % = 1/5
Also we have,
P(E/A) = 2 % = 1/50
P(E/B) = 2 % = 1/50
P(E/C) = 3 % = 3/100
Now,
P(the defective item drawn was manufactured on machine A) =
=
=
=
=
= 500/1100
= 5/11
Solution:
Let us assume
A = the event of choosing two-headed coin,
B = the event of choosing a biased coin that comes up head 75 % of the times
C = the event of choosing a biased coin that comes up tail 40 % of the times
E = the event of getting a head.
Now,
P(A) = 1/3
P(B) = 1/3
P(B) = 1/3
Also we have,
P(E/A) = 1
P(E/B) = 75 % = 75/100 = 3/4
P(E/C) = 60 % = 60/100 = 3/5
By using Bayes' theorem, the required probability is
P(the head shown was of two-headed coin) = P (A/E)
=
=
=
=
=
= 20/47
Solution:
Let us assume that the events are
A = the item is defective
E1 = machine A is chosen
E2 = machine B is chosen
E3 = machine C is chosen
So,
P(E1) = 30/100
P(E2) = 25/100
P(E3) = 45/100
Now,
P(A/E1) = 1/100
P(A/E2) = 1.2/1000
P(A/E3) = 2/100
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 30/150
= 1/5
= 0.2
Solution:
Let us assume that the events are
A = the cycle is of standard quality
E1 = plant I is chosen
E2 = plant II is chosen.
So,
P(E1) = 60/100
P(E2) = 40/100
Now,
P(A/E1) = 80/100
P(A/E2) = 90/100
By using Bayes' theorem, the required probability is
P(E2/A) =
=
=
= 36/84
= 3/7
Solution:
Let us assume that the events are
A = the ball is red
E1 = urn A is chosen
E2 = urn B is chosen
E2 = urn C is chosen.
So, P(E1) = 1/3
P(E2) = 1/3
P(E3) = 1/3
Now,
P(A/E1) = 6/10 = 3/5
P(A/E2) = 2/8 = 1/4
P(A/E3) = 1/6
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 36/61
Solution:
Let us assume that the events are
A = the person suffers from the disease
E1 = a smoker and a non-vegetarian
E2 = a smoker and a vegetarian
E3 = a non-smoker and a vegetarian
So, P(E1) = 160/400
P(E2) = 100/400
P(E3) = 140/400
Now,
P(A/E1) = 35/100
P(A/E2) = 20/100
P(A/E3) = 10/100
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 560/900
= 28/45
Solution:
Let us assume that the events are
A = the item is defective
E1 = machine A is chosen
E2 = machine B is chosen
E3 = machine C is chosen
So, P(E1) = 100/600
P(E2) = 200/600
P(E3) = 300/600
Now,
P(A/E1) = 2/100
P(A/E2) = 3/100
P(A/E3) = 5/100
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 2/23
Solution:
Let us assume the events are
A = the ball is white
E1 = bag I is chosen
E2 = bag II is chosen
So, P(E1) = 1/2
P(E2) = 1/2
Now,
P(A/E1) = 1/7
P(A/E2) = 4/7
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 1/5
Solution:
Let us assume the events are
A = The height of the student is more than 1.75 m
E1 = The selected student is a girl
E2 = The selected student is a boy
So,
P(E1) = 60/100
P(E2) = 40/100
Now,
P(A/E1) = 1/100
P(A/E1) = 4/100
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 6/22
= 3/11
Solution:
Let us assume the events are
A = The change takes place
E1 = A is selected
E2 = B is selected
E3 = C is selected
So,
P(E1) = 4/7
P(E2) = 1/7
P(E3) = 2/7
Now,
P(A/E1) = 0.3
P(A/E2) = 0.8
P(A/E3) = 0.5
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 1.2/3
= 2/5
1 - P(E1/A) = 1 - 2/5 = 3/5
Solution:
Let us assume the events are
E1 = The selection of A as manager
E2 = The selection of B as manager
E3 = The selection of C as manager
So,
P(E1) = The probability of selection of A = 1/7
P(E2) = The probability of selection of B = 2/7
P(E3) = The probability of selection of C = 4/7
Let us assume that A be the event representing the change not taking place.
P(A/E1) = Probability that A does not introduce change = 0.2
P(A/E2) = Probability that B does not introduce change = 0.5
P(A/E3) = Probability that C does not introduce change = 0.7
So, the required probability = P(A/E3)
By using Bayes' theorem, the required probability is
P(A/E3) =
=
=
= 2.8/4
= 0.7
Solution:
Let us assume the events are
A = The vehicle meets the accident
E1 = is a scooter
E2 = is a motorcycle
So,
P(E1) = 2000/5000 = 0.4
P(E2) = 3000/5000 = 0.6
Now,
P(A/E1) = 0.01
P(A/E2) = 0.02
By using Bayes' theorem, the required probability is
P(E2/A) =
=
=
= 0.012/0.016
= 3/4
Solution:
Let us assume the events are
A = The selected student attains grade A
E1 = resides in a hostel
E2 = does not reside in a hostel
So, P(E1) = 60/100
P(E2) = 40/100
Now,
P(A/E1) = 30/100
P(A/E2) = 20/100
By using Bayes' theorem, the required probability is
P(E1/A) =
=
=
= 9/13
This chapter covers advanced probability concepts, building upon the fundamental definitions and principles introduced in the earlier chapters. It delves into topics such as conditional probability, the multiplication principle, the law of total probability, Bayes' theorem, and probability distributions. The chapter presents various examples and problem-solving techniques to help students develop a strong understanding of these concepts and their applications in real-world scenarios. Key emphasis is placed on analyzing complex probability problems, formulating appropriate mathematical models, and deriving accurate solutions.
1. Two coins are tossed simultaneously. Find the probability of getting at least one head.
2. In a family, the probability of having a boy is 0.52 and the probability of having a girl is 0.48. Find the probability of having at least one boy in a family with two children.
3. A bag contains 5 red, 3 blue, and 4 green balls. Two balls are drawn at random without replacement. Find the probability that both balls are of different colors.
4. A and B are two events such that P(A) = 0.6, P(B) = 0.4, and P(A and B) = 0.2. Find P(A|B) and P(B|A).
5. The probability of a student passing in Mathematics, Physics, and Chemistry are 0.8, 0.7, and 0.6, respectively. If the probability of passing in at least one subject is 0.95, find the probability of passing in all three subjects.
6. In a certain city, the probability of rain on a given day is 0.3. If it rains on a particular day, the probability of a traffic jam is 0.6. If it does not rain, the probability of a traffic jam is 0.2. Find the probability of a traffic jam.
7. A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a king or a spade.
8. A company manufactures light bulbs, of which 5% are defective. A sample of 10 light bulbs is randomly selected. Find the probability that exactly 2 of the bulbs are defective.
9. In a certain college, 60% of the students are female, and 40% of the female students are good in mathematics. Find the probability that a randomly selected student is good in mathematics, given that the student is female.
10. A bag contains 5 white balls and 3 black balls. Two balls are drawn at random without replacement. Find the probability that the second ball drawn is white.