NCERT Solutions for Class 12 Maths, Chapter 13, Exercise 13.3, have been created by GeeksforGeeks to assist students in their academic journey. There are a total of 14 questions in Exercise 13.3 of Chapter 13 in the Class 12 Maths NCERT textbook. In this article, we will demonstrate how to solve all the problems in this exercise. Students can benefit from these error-free solutions to achieve the best results.
Chapter 13: Probability - Exercise 13.3
Question 1: An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Solution:
Let B : first ball drawn is black
R : second ball is red
Probability that the second ball is red = Probability that the first ball is black × Probability that the second ball is red if first is black + Probability that first ball is red × Probability that second ball is red if first is red.
Probability first ball is black and second ball is red
Probability first ball is black = 5/(5 + 5) = 5/10 = 1/2
Now, there are 5 Red and (5 + 2 =) 7 Black balls in the urn. So, Probability second ball is red, when first ball is black = 5/(5 +7) = 5/12
Probability first ball is red and second ball is red
Probability first ball is red = 1 - P(first is black) = 1 - 1/2 = 1/2
Now, there are (5 + 2 =) 7 Red and 5 Black balls in the urn
So, Probability second ball is red, when first ball is red = 7/(5 + 7) = 7/12
Thus, Probability that the second ball is red = Probability that the first ball is black × Probability that the second ball is red if first is black + Probability that first ball is red × Probability that second ball is red if first is red.
⇒ Probability that the second ball is red = 1/2 × 5/12 + 1/2 × 7/12
⇒ Probability that the second ball is red = 5/24 + 7/24 = 12/24 = 1/2
Therefore, Probability that the second ball is red is 1/2
Question 2: A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.4 Red (R) 4 Black (B) 2 Red (R) 6 Black (B) Let B1 : ball is drawn from Bag I B2 : ball is drawn from Bag II R : ball is drawn is red
Solution:
We need to find Probability that ball is drawn from Bag I, if ball is red = P(B1|R)
So, P(B1|R) = (P(B1) . P(R|B1))/(P(B1) . P(R|B1)+P(B2) . P(R|B2))
P(B1) = Probability that ball is drawn from Bag I = 1/2
P(R|B1) = Probability that ball is red, if drawn from Bag I = 4/(4 + 4) = 4/8 = 1/2
P(B2) = Probability that ball is drawn from Bag II = 1/2
P(R|B2) = Probability that ball is red, if drawn from Bag II = 3/(3 + 5) = 3/8 = 1/4
Putting values in formula, P(B1|R) = (1/2 × 1/2)/(1/2 × 1/2 + 1/2 × 1/4) = (1/4)/(1/4 + 1/8) = (2/8)/(3/8) = 2/3
Therefore, required probability is 2/3
Question 3: Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostler?
Solution:
Let H : student selected is a hostler
D : student selected is a day scholar
A : student has an 'A' grade
We need to find the Probability that the student selected is a hostler, if he has an 'A' grade. i.e. P(H|A)
So, P(H|A) = P(H) . P(A|H)/ P(D) . P(A|D) + P(H) . P(A|H)
P(H) = Probability that student a hostler = 60% = 60/100 = 0.6
P(A|H) = Probability that student gets 'A' grade, if hostler = 30% = 30/100 = 0.3
P(D) = Probability that student a day scholar = 40% = 40/100 = 0.4
P(A|D) = Probability that student gets 'A' grade, if day scholar = 20% = 20/100 = 0.2
Putting values in formula, P(H|A) = (0.6 × 0.3)/(0.4 × 0.2 + 0.6 × 0.3)
= (0.18)/(0.08 + 0.18) = (0.18)/(0.26) = 18/26 = 9/13
Therefore, required probability is 9/13
Question 4: In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?Let A : student know the answer B : student guesses C : student answers correctly.
Solution:
We need to find the Probability that the student knows the answer, if he answered it correctly i.e. P(A|C)
P(A|C) = (P(A) . P(C|A))/P(A) . P(C|A) + P(B) P(C|B)
P(A) = Probability that student knows the answer = 3/4
P(C|A) = Probability that the student answered correctly, if he knows the answer = 1
P(B) = Probability that the student guesses the answer = 1/4
P(C|B) = Probability that the student answered correctly, if he guesses = 1/4
Putting values in formula, P(A|C) = (3/4 × 1) / (1/4 × 1/4 + 3/4 × 1) = (3/4) / (1/16 + 3/4) = (3/4) / (13/16) = (12/16) / (13/16) = 12/13
Therefore, required probability is 12/13
Question 5: A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution:
Let A : Person has the disease
B : Person does not have the disease
C : Test result is positive
We need to find the Probability that a person has the disease given that his test result is positive i.e. P(A|C)
P(A|C) = (P(A).P(C|A))/P(B) . P(C|B) + P(A) P(C|A)
P(A) = Probability that person has disease = 0.1% = (0.1)/100 = 0.001
P(C|A) = Probability that test result is positive, if the person has the disease = 99% = 99/100 = 0.99
P(B) = Probability that person does not have the disease = 99.9% = (99.9)/100 = 0.999
P(C|B) = Probability that test result is positive, if the person does not have the disease = 0.005
Putting values in formula, P(A|C) = (0.001 × 0.99) / (0.999 × 0.005 + 0.001 × 0.99)
= (99 × 10-5) / (499.5 × 10-5 + 0.001 × 0.99)
= (10-5 × 99) / (10-5) × [499.5 + 99])
= 99/(598.5) = 990/5985 = 22/133
Therefore, required probability is 22/133
Question 6: There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Solution:
Let C1: two headed coin
C2: biased coin
C3: unbiased coin
H : head appears on the coin
We need to find Probability that coin is two headed, if it shows head i.e. P(C1|H)
P(C1|H) = (P(C1).P(H|C1))/(P(C1).P(H|C1)+P(C2).P(H|C2)+P(C3).P(H|C3))
P(C1) = Probability that coin selected is two headed = 1/3
P(H|C1) = Probability that head appear on the coin C1 = 1
P(C2) = Probability that coin selected is biased = 1/3
P(H|C2) = Probability that head appear on the coin C2 = 75% = 75/100 = 3/4
P(C3) = Probability that coin selected is unbiased = 1/3
P(H|C3) = Probability that head appear on the coin C3 = 1/2
Putting values in formula, P(C1|H) = (1/3 × 1) / (1/3 × 1 + 1/3 × 3/4 + 1/3 × 1/2)
⇒ P(C1|H) = (1/3 × 1) / (1/3 [1 + 3/4 + 1/2])
⇒ P(C1|H) = 1 / (9/4)
⇒ P(C1|H) = 4/9
Therefore, required probability is 4/9
Question 7: An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution:
Let S : Scooter driver met with accident
C : Car driver met with accident
T : Truck driver met with accident
A : the driver is insured
We need to find the Probability that the person met with an accident is a scooter driver, if he is insured i.e. P(S|A)
P(S|A) = (P(S) . P(A|S))/(P(S) . P(A|S)+P(C) . P(A|C)+P(T) . P(A|T))
P(S) = Probability that a scooter driver met with an accident = 0.01
P(A|S) = Probability that a scooter driver is insured = 2000/(2000 + 4000 + 6000) = 2000/12000 = 1/6
P(C) = Probability that a car driver met with an accident = 0.03
P(A|C) = Probability that a car driver is insured = 4000/(2000 + 4000 + 6000) = 4000/12000 = 1/3
P(T) = Probability that a Truck driver met with an accident = 0.15
P(A|T) = Probability that a Truck driver is insured = 6000/(2000 + 4000 + 6000) = 6000/12000 = 1/2
Putting values in formula, P(S|A) = (0.01 × 1/6) / (0.01 × 1/6 + 0.03 × 1/3 + 0.15 × 1/2)
⇒ P(S|A) = (0.01 × 1/6) / (0.01 [1/6 + 3 × 1/3 + 15 × 1/2])
⇒ P(S|A) = (1/6) / (1/6 + 1 + 15/2)
⇒ P(S|A) = (1/6) / (52/6)
⇒ P(S|A) = 1/52
Therefore, required probability is 1/52
Question 8: A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Solution:
Let A : Items produced by machine A
B : Items produced by machine B
D : Items is defective
We need to find the Probability that the item was produced by machine B, if it is found to be defective i.e. P(B|D)
P(B|D) = (P(B) . P(D|B)) / (P(B) . P(D|B) + P(A) . P(D|A))
P(A) = Probability that the item is produced by machine A = 60% = 60/100 = 0.6
P(D|A) = Probability that the item is defective, if produced by machine A = 2% = 2/100 = 0.02
P(B) = Probability that the item is produced by machine B = 40% = 40/100 = 0.4
P(D|B) = Probability that the item is defective, if produced by machine B = 1% = 1/100 = 0.01
Putting values in formula, P(B|D) = (0.4 × 0.01) / (0.6 × 0.02 + 0.4 × 0.01)
⇒ P(B|D) = (0.01 × 0.4) / (0.01 [0.6 × 2 + 0.4])
⇒ P(B|D) = 0.4 / (1.2 + 0.4)
⇒ P(B|D) = 0.4 / 1.6
⇒ P(B|D) = 4/16 = 1/4
Therefore, required probability is 1/4
Question 9: Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Solution :
Let E : first group wins
F : second group wins
G : new product is introduced
We need to find the Probability that the new product introduced was by the second group i.e. P(F|G)
P(F|G) = (P(F). P(G|F))/(P(E). P(G|E) + P(F). P(G|F))
P(E) = Probability that first group wins = 0.6
P(G|E) = Probability that new product is introduced, if first group wins = 0.7
P(F) = Probability that the second group wins = 0.4
P(G|F) = Probability that a new product is introduced, if second group wins = 0.3
Putting values in formula, P(F|G) = (0.4 × 0.3)/(0.6 × 0.7 + 0.4 × 0.3)
⇒ P(F|G) = (0.12)/(0.42 + 0.12)
⇒ P(F|G) = (0.12)/(0.54) = 2/9
Therefore, required probability is 2/9
Question 10: Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Solution:
Let A : 1, 2, 3, 4 appear on the die
B : 5, 6 appear on the die
C : exactly one head is obtained
We need to find the Probability that if exactly one head is obtained on the toss of a coin, she threw 1, 2, 3 or 4 with the die i.e. P(A|C)
P(A|C) = (P(A) . P(C|A))/(P(A) . P(C|A) + P(B) . P(C|B))
P(A) = Probability that 1, 2, 3 or 4 appear on the die = 4/6 = 2/3
P(C|A) = Probability that exactly one head is obtained, if 1, 2, 3 or 4 appear on the die = 1/2
P(B) = Probability that 5 or 6 appear on the die = 2/6 = 1/3
P(C|B) = Probability that exactly one head is obtained, if 5 or 6 appear on the die = 3/8
Putting values in formula, P(A|C) = (2/3 × 1/2) / (2/3 × 1/2 + 1/3 × 3/8)
⇒ P(A|C) = (1/3) / (1/3 [1 + 3/4])
⇒ P(A|C) = (1/3) / (1/3 × 7/4)
⇒ P(A|C) = 2 / (2 + 3/4)
⇒ P(A|C) = 2 / (11/4) = 8/11
Therefore, required probability is 8/11.
Question 11: A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Solution:
Let A : Event that item produced by operator A
B : Event that item produced by operator B
C : Event that item produced by operator C
D : Event that item produced is defective
We need to find out the Probability that item is produced by operator A if it is defective i.e. P(A|D)
P(A|D) = (P(A) . P(D|A)) / (P(A) . P(D|A) + P(B) . P(D|B) + P(C) . P(D|C))
P(A) = Probability of item is produced by operator A = 50% = 50/100 = 0.5
P(D|A) = Probability of a defective item produced by operator A = 1% = 1/100 = 0.01
P(B) = Probability of item is produced by operator B = 30% = 30/100 = 0.3
P(D|B) = Probability of a defective item produced by operator B = 5% = 5/100 = 0.05
P(C) = Probability of item is produced by operator C = 20% = 20/100 = 0.2
P(D|C) = Probability of a defective item produced by operator C = 7% = 7/100 = 0.07
Putting Values in the formula
P(A|D) = (0.5 × 0.01) / (0.5 × 0.01 + 0.3 × 0.05 + 0.2 × 0.07)
⇒ P(A|D) = 0.005 / (0.005 + 0.015 + 0.014)
⇒ P(A|D) = 0.005 / 0.034 = 5/34
Therefore, required probability = 5/34
Question 12: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Solution:
Let, E1 : Event that lost card is diamond
E2 : Event that lost card is not a diamond
A : Event that two cards drawn are diamond
We need to find out the probability that the lost card being a diamond if two cards drawn are found to be both diamond. i.e. P(E1|A)
P(E1|A) = (P(E1).P(A|E1))/(P(E1).P(A|E1)+P(E2).P(A|E2))
P(E1) = Probability that lost card is diamond = 13/52 = 1/4
P(A|E1) = Probability of getting 2 diamond cards if lost card is diamond
= (Selecting 2 diamond cards from (13-1=12) diamond cards)/(Selecting any 2 cards from 51 cards)
= 12C2/51C2
= ((12 × 11)/2!)/((51 × 50)/2!)
= (12 × 11)/(51 × 50)
P(E2) = Probability that lost card is not a diamond = 1 - P(E1) = 1 - 1/4 = 3/4
P(A|E2) = Probability of getting 2 diamond cards if lost card is not a diamond
= (Selecting 2 diamond cards from 13 diamond cards)/(Selecting any 2 cards from 51 cards)
= 13C2/51C2
= ((13 × 12)/2!)/((51 × 50)/2!)
= (13 × 12)/(51 × 50)
Putting value in the formula,
P(E1|A) = (P(E1).P(A|E1))/(P(E1).P(A|E1)+P(E2).P(A|E2))
⇒ P(E1|A) = (1/4 × (12 × 11)/(51 × 50))/(1/4 × (12 × 11)/(51 × 50) + 3/4 × (13 × 12)/(51 × 50))
⇒ P(E1|A) = (1/4 × (12)/(51 × 50) × 11)/(1/4 × 12/(51 × 50) × (11 + 3 × 13))
⇒ P(E1|A) = 11/(11 + 39) = 11/50
Therefore, required probability is 11/50
Question 13: Probability that A speaks truth is 4/5 . A coin is tossed. A reports that a head appears. The probability that actually there was head is
(A) 4/5 (B) 1/2 (C) 1/5 (D) 2/5
Let E : A speaks truth
F : A Lies
H : head appears on the toss of a coin
We need to find the Probability that head actually appears, if A reports that a head appears i.e. P(E|H)
P(E|H) = "P(E) . P(H|E)" / "P(F) . P(H|F) + P(E) . P(H|E)"
P(E) = Probability that A speaks truth = 4/5
P(H|E) = Probability that head appears, if A speaks truth = 1/2
P(F) = Probability that A lies = 1 – P(E) = 1 – 4/5 = 1/5
P(H|F) = Probability that head appears, if A lies = 1/2
Putting values in formula,
P(E|H) = (4/5 × 1/2)/(1/5 × 1/2 + 4/5 × 1/2) = (1/5 × 1/2 × 4)/(1/5 × 1/2 [1 + 4]) = 4/5
Therefore, required probability is 4/5
∴ A is the correct answer
Question 14: If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
(A) P(A|B) = P(B)/P(A)
(B) P(A| B) < P(A)
(C) P(A|B) ≥ P(A)
(D) None of these
Solution:
Given, A ⊂ B
So, A ∩ B = A
P(A ∩ B) = P(A)
Now, P(A|B) = (P(A ∩ B))/(P(B))
P(B) = (P(A ∩ B))/(P(A|B))
P(B) = (P(A))/(P(A|B))
Also, 0 < P(B) ≤ 1
⇒ 0 < (P(A))/(P(A|B)) ≤ 1
⇒ 0 × P(A|B) < (P(A))/(P(A|B)) × P(A|B) ≤ 1 × P(A|B)
⇒ 0 < P(A) ≤ P(A|B)
∴ P(A|B) ≥ P(A)
Hence, Option (C) is Correct
Summary
Exercise 13.3 focuses on the concept of random variables and probability distributions. It introduces discrete random variables, their probability mass functions (PMF), and cumulative distribution functions (CDF). Students learn to calculate expected values, variances, and standard deviations of discrete random variables. The exercise also covers properties of probability distributions and how to solve problems involving real-life applications of random variables.
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