Author name: Prasanna

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-5/

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.5

Question 1.
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
(i) 5 successes?
(ii) at least 5 successes?
(iii) at most 5 successes?
Solution:

Question 2.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
Solution:
This is a Bernoulli trial with n = 4
Here success is getting a doublet
p = probability of success
= \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
q = 1 – p = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
Let X be the number of successes. Then X is a binomial distribution with

Question 3.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability, of two successes.
Solution:
This is a Bernoulli trial with n = 10
Here success is the inclusion of a defective item
p = probability of success = \(\frac { 5 }{ 100 }\) = \(\frac { 1 }{ 20 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 20 }\) = \(\frac { 19 }{ 20 }\)
Let X be the number of defective item in the sample.
Then X is a binomial distribution with

Question 4.
Five cards are drawn successively with replacement from a well- shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is spade?
Solution:
This is a Bernoulli trial with n = 5
Here success is getting a spade
p = probability of success = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 4 }\) = \(\frac { 3 }{ 4 }\)
Let X denote the number of spade cards.
Then X is a binomial distribution with
n = 5, p = \(\frac { 1 }{ 4 }\) and q = \(\frac { 3 }{ 4 }\)
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i. Probability of 5 spades = P(X = 5)

Question 5.
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use
Solution:
This is a Bernoulli trial with n = 5
Here success is the bulb fuse after 150 days.
∴ p = probability of success = 0.05
∴ q = 1 – p = 1 – 0.05 = 0.95
Let X be the number of bulbs fuse after 150 days
Then X is a binomial distribution with n = 5, p = 0.05 and q = 0.95
∴ P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) When X = 0, P(X = 0)
\({ }^{5} \mathrm{C}_{0}(0.05)^{0}(0.95)^{5-0}=(0.95)^{5}\)

(ii) X is not more than 1 i.e. X ≤ 1
∴ P(X ≤ 1) = P(X = 0) + P(X = 1)
= \({ }^{5} \mathrm{C}_{0}(0.05)^{0}(0.95)^{5}+{ }^{5} \mathrm{C}_{1}(0.05)(0.95)^{4}\)
= (0.95)⁴ [0.95 + 5 x 0.05]
= (0.95)⁴ x 1.2

(iii) X is more than 1 i.e. X > 1
P(X > 1) = 1 – P(X ≤ 1)
= 1 – (0.95)⁴ x 1.2 from (ii)

(iv) X is atleast 1 i.e. X ≥ 1
P(X ≥ 1) = 1 – P(X = 0)
= 1 – (0.95)⁵ from (i)

Question 6.
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four bails are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Solution:
This is a Bernoulli trial with n = 4.
Here success is the ball drawn is marked with number 0
p = probability of success = \(\frac { 1 }{ 10 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 10 }\) = \(\frac { 9 }{ 10 }\)
Let X be the number of balls marked with 0
Then X is a binomial distribution with

In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls head, he answers ‘true’, if it falls tail, he answers ‘false’. Find

Question 7.
In an examination, 20 questions of true – false type are asked. Suppose a student tosses fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true,’ if it falls tails, he answers “ false’. Find the probability that he answers at least 12 questions correctly.
Solution:
Probability that student answers a question true = \(\frac { 1 }{ 2 }\)
i.e., when a coin is thrown, probability that a head is obtained = \(\frac { 1 }{ 2 }\)
Probability that his answer is false = \(1-\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
Probability that his answer at least 12 questions correctly = P (12) + P (13) + P (14) +…….. P (20)

Question 8
Suppose X has a binomial distribution \(B\left( 6,\frac { 1 }{ 2 } \right) \). Show that X = 3 is the most likely outcome.
(Hint: P (X = 3) is the maximum among all P (Xi), xi. = 0,1,2,3,4,5,6)
Solution:
\({ \left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) }^{ 6 } \)

Question 9.
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Solution:
P = \(\frac { 1 }{ 3 }\). q = 1 – P = \(1-\frac { 1 }{ 3 }\) = \(\frac { 2 }{ 3 }\)

Question 10.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \(\frac { 1 }{ 100 }\) . What is the probability that he will win a prize?
(a) at least once,
(b) exactly once,
(c) at least twice?
Solution:
Probability that the person wins the prize = \(\frac { 1 }{ 100 }\)
Probability of losing = \(1-\frac { 1 }{ 100 }\) = \(\frac { 99 }{ 100 }\)
(a) Probability that he loses in all the loteries

Question 11.
Find the probability of getting 5 exactly twice in 7 throws of a die.
Solution:
S = {1,2,3,4,5,6},n(S) = 6
A = {5} ⇒ n(A) = 1

Question 12.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
When a die is thrown,
Probabiltiy of getting a six = \(\frac { 1 }{ 6 }\)
Probabiltiy of not getting a six = \(1-\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
Probabiltiy of getting at most 2 sixes in 6 throws of a single die = P (0) + P (1) + P (2)

Question 13.
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles 9 are defective?
Solution:
This is a Bernoulli trial with n = 12
Here success is selecting a defective item.
p = probability of success = \(\frac { 1 }{ 100 }\) = \(\frac { 1 }{ 10 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 10 }\) = \(\frac { 9 }{ 10 }\)
Let X is the number of defective items. Then X is a binomial distribution with

Question 14.
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(a) \({ 10 }^{ -1 }\)
(b) \({ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)
(c) \({ \left( \frac { 9 }{ 10 } \right) }^{ 5 }\)
(d) \(\frac { 9 }{ 10 }\)
Solution:
p = \(\frac { 1 }{ 10 }\)
q = \(\frac { 9 }{ 10 }\)
n = 5,
r = 0,
P(X = 0) = \({ \left( \frac { 9 }{ 10 } \right) }^{ 5 }\)
Option (c) is correct

Question 15.
The probability that a student is not a swimmer is \(\frac { 1 }{ 5 }\). Then the probability that out of five students, four are swimmers is:
(a) \({ }^{5} \mathrm{C}_{4}\left(\frac{4}{5}\right)^{4} \frac{1}{5}\)
(b) \(\left(\frac{4}{5}\right)^{4} \frac{1}{5}\)
(c) \({ }^{5} C_{1} \frac{1}{5}\left(\frac{4}{5}\right)^{4}\)
(d) None of these
Solution:
This is Bernoulli trial with n = 5
Here success is that student is a swimmer.
p = probability of success
= 1 – P (not a swimmer) = 1 – \(\frac { 1 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
∴ q = 1 – p = 1 – \(\frac { 4 }{ 5 }\) = \(\frac { 1 }{ 5 }\)
Let X denote the number of swimmers in the group of 5 students
Then X is a binomial distribution with

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-3/

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.3

Exercise 13.3 Class 12 NCERT Solutions Question 1.
An urn contain 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:
Urn contain 5 red and 5 black balls.
(i) Let a red ball is drawn.
probability of drawing a red ball = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\)
Now two red balls are added to the urn.
⇒ The urn contains 7 red and 5 black balls.
Probability of drawing a red ball = \(\frac { 7 }{ 12 }\)

(ii) Let a black ball is drawn at first attempt
Probability of drawing a black ball = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\)
Next two black balls are added to the urn
Now urn contains 5 red and 7 black balls
Probability of getting a red ball = \(\frac { 5 }{ 12 }\)
⇒ Probability of drawing a second ball as red
= \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } +\frac { 1 }{ 2 } \times \frac { 5 }{ 12 } =\frac { 7 }{ 24 } +\frac { 5 }{ 24 } =\frac { 12 }{ 24 } =\frac { 1 }{ 2 }\)

Ex 13.3 Class 12 Maths Ncert Solutions 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.
Solution:
E₁ : first bag is selected
E₂ : second
bag is selected A : ball drawn is red
E₁ and E₂ are mutually exclusive & exhaustive events
P(E₁) = P(E₂) =
P(A|E₁) = \(\frac { 4 }{ 8 }\) and P(A|E₂) = \(\frac { 2 }{ 8 }\)
By Bayes’ theorem

Class 12 Maths Chapter 13 Ex 13.3 NCERT Solutions 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 hostlier?
Solution:
E₁ : student is a hosteler
E₂ : student is a day scholar
A : student attains A grade.
E₁ and E₂ are mutually exclusive and exhaustive

Question 4.
In answering a question on a multiple choice test, a student either knows the answer or 3 guesses. Let \(\frac { 3 }{ 4 }\) be the probability that he knows the answer and \(\frac { 1 }{ 4 }\) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \(\frac { 1 }{ 4 }\) . What is the probability that the student knows the answer given that he answered it correctly?
Solution:
Let E₁ : student knows the answer
E₂ : student guesses the answer
A : student answers correctly
E₁ and E₂ are mutually exclusive and exhaustive events.

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 E₁ : the person has a disease
E₂ : the person is healthy
A : the test result is positive
E₁ and E₂ are mutually exclusive and exhaustive events.

Question 6.
There are three coins. One is a two headed coin, another is a biased coin that conies up heads 75% of the time and third is an unbiased coin. One of the three coins is choosen at random and tossed, it shows head, what is the probability that it was the two headed coin?
Solution:
Let
E₁ : coin tossed is two headed
E₂ : coin tossed is biased
E₃ : coin tossed is unbiased
A: getting head
E₁ and E₂ and E₃ are mutually exclusive and exhaustive events.

Question 7.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of mi accident are 0.01, 0.03, 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution:
Let
E₁ : insured person is a scooter driver
E₂ : insured person is a car driver
E₃ : insured person is a truck driver and
A : insured person meets with an accident.
E₁ E₂ and E₃ are mutually exclusive and exhaustive events.
Total number of insured person
= 2000 + 4000 + 6000 = 12000

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
E₁ : item produced by machine A
E₁ : item produced by machine B
A : item is defective
E₁ and E₂ are mutually exclusive and exhaustive events

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 E₁ : first group wins
E₂ : second group wins
A : introducing a new product
E₁ and E₂ are mutually exclusive and exhaustive events

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:
When she gets 5,6, she throws a coin three times.
The events are
E₁ : getting 1, 2, 3, 4
E₂ : getting 5, 6
A : getting exactly one head
E₁ and E₂ are mutually exclusive and exhaustive events
P(E₁) = \(\frac { 4 }{ 6 }\) P(E₂) = \(\frac { 2 }{ 6 }\)
When she gets 1,2,3,4, .she throws a coin once.
P(A|E₁) = \(\frac { 1 }{ 2 }\)
When a coin is tossed 3 times, the sample space is {HHH, HHT,- HTH, HTT, THH, THT, TTH, TTT}
One head is obtained as {HTT, THT, TTH} P(A|E₂) = \(\frac { 3 }{ 8 }\)

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 E₁ : item produced by operator A
E₂ : item produced by operator B
E₃ : item produced by operator C
A : event of getting a defective item
E₁, E₂ and E₃ are pairwise disjoint and exhaustive events

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: lost card is a diamond
E₂ : lost card is not a diamond
A : getting 2 diamond cards from 51 cards
E₁, E₂ are mutually exclusive and exhaustive events

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) \(\frac { 4 }{ 5 }\)
(b) \(\frac { 1 }{ 2 }\)
(c) \(\frac { 1 }{ 5 }\)
(d) \(\frac { 2 }{ 5 }\)
Solution:
Let E₁ getting a head
E₂: getting a tail
F : A reports ‘head occurred’
E₁ and E₂ are mutually exclusive and ex-haustive events

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) = \(\frac { P(B) }{ P(A) }\)
(b) P (A | B) < P (A)
(c) P(A | B) ≥ P(A)
(d) None of these
Solution:

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise

Question 1.
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and.atmost 300 units of cholesterol. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maxi¬mum amount of vitamin A in the diet?
Solution:
Let x and y be the number of packets of food P and Q. We have the data as

Calcium constraint: 12x + 3y ≥ 240
i.e., 4x + y ≥ 80
Iron constraint: 4x + 20y ≥ 460
i.e., x + 5y ≥ 115
Cholesterol constraint: 6x + 4y ≤ 300
i.e., 3x + 2y ≤ 150
Quantity of vitamin A in the diet Z = 6x + 3y
The LPP is Minimise Z = 6x + 3y
subject to the constraints
4x + y ≥ 80
x + 5y ≥ 115
3x + 2y ≤ 150
x ≥ 0, y ≥ 0

The feasible region is shaded in the figure. We use corner point method to find the minimum value of Z.

Minimum value of Z is at B (15,20). Hence, the amount of vitamin A under constraints given in the problem will be minimum, if 15 packets of food P and 20 packets of food Q are used in the special diet. The minimum quantity of vitamin A will be 15 0 units.
∴ Maximum value of Z is 285 at C(40, 15).
Hence 40 packets of food P and 15 packets of food Q should be used to maximise the amount of vitamin A. The maximum amount of vitamin A in the diet is 285 units.

Question 2.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing ₹ 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹ 200 per bag, contains 1.5 units of nutritional element A, 1.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Solution:
Let x and y denote the number of bags of brand P and brand Q respectively.

Element A constraint: 3x +1, 5y ≥ 18
Element B constraint: 2.5x + 11.25y ≥ 45
Element C constraint: 2x + 3y ≥ 24
Cost function : Z = 250x + 200y
The L.P.P is minimise Z = 250x + 200y
subject to the constraints 3x + 1.5y ≥ 18, 2.5x + 11.25y ≥ 45, 2x + 3y ≥ 24, x, y ≥ 0.

The feasible region is shaded in the figure.

From the table the minimum value of Z is 1950. Since the region is unbounded 1950 may or may not be the minimum value of Z. Consider the inequality 250x + 200y < 1950. This half plane has no point common with the feasible region The minimum value of Z is 1950 at C(3, 6). To produce a mixture having minimum cost, the farmer should mix 3 bags’ of brand P and 6 bags of brand Q.
The minimum cost of the mixture is ₹ 1950

Question 3.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

One kg of food X costs Rs. 16 and one kg of food Y costs Rs.20. Find the least cost of the mixture which will produce the required diet.
Solution:
Let x kg of food X and y kg of food Y be mixed in the diet. We have the data as

Vitamin A constraint: x + 2y ≥ 10
Vitamin B constraint: 2x + 2y ≥ 12
Vitamin C constraint: 3x + y ≥ 18
Cost function Z = 16x + 20y
The L.P.P. is
Minimise Z = 16x + 20y
subject to the constraints
x + 2y ≥ 10, 2x + 2y ≥ 12, 3x + y ≥ 8, x, y ≥ 0

From the table, the minimum value of Z is 112. Since the region is unbounded 112 may or may not be the minimum value of Z. Consider the inequality 16x + 20y =112. This half plane has no point common with the feasible region.
∴ Minimum value of Z is 112 at B(2, 4). Hence 2kg of food X and 4kg of food Y should be mixed. The least cost of the mix-ture is ₹ 112

Question 4.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given be-low:

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹ 7.50 and that on each toy of type B is ₹ 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Solution:
Let x be the no. of toys of type A and y be the no. of toys of type B.
We have the data as

Machine I constraint: 12x + 6y ≤ 360
Machine II constraint: 18x ≤ 360
Machine II constraint: 6x + 9y ≤ 360
Profit function : Z = 7.5 x + 5y
The LPP is maximise Z = 7.5 x + 5y
subject to the constraints 12x + 6y ≤ 360, 18x + 0y ≤ 360, 6x + 9y ≤ 360, x, y ≥ 0.

∴ Maximum value of Z is 262.50 at C(15, 30).
Hence 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit.

Question 5.
An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Solution:
Let x tickets of executive class and y tickets of economy class be sold.
We have the data as

Total ticket constraint: x + y ≤ 200
Executive ticket constraint: x ≥ 20
Economy ticket constraint: y ≥ Ax
Profit function : Z = 1000x + 600y
The L.P.P. is maximise Z = 1000x + 600y
subject to the constraints x + y ≤ 200, x ≥ 20, y ≥ 4x, x, y ≥ 0.

The feasible region is shaded in the figure.

Hence 40 tickets of executive class and 160 tickets of economy class are to be sold to get the maximum profit. The maximum profit is ₹ 136000

Question 6.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60,50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let x quintals of grains are transported from godown Ato ration shop D andy quintals be transported from godown Ato ration shop E.
We have the data as

The transported quantity of rice is non negative x ≥ 0, y ≥ 0, 100 – (x + y) ≥ 0, 60 – x ≥ 0
50 – y ≥ 0 , x + y – 60 ≥ 0
i.e., x + y ≤ 100, x + y ≥ 60, x ≤ 60, y ≤ 50, x ≥ 0, y ≥ 0
The total transportation cost
Z = 6x + 3y + 2.5 (100 – (x + y) + 4(60 – x) + 2(50 – y) + 3 (x + y – 60)
Z = 2.5x + 1.5y + 410
The L.P.P. is Minimise Z = 2.5x + 1.5y + 410
subject to the constraints x ≤ 60, y ≤ 50, x + y ≥ 60, x + y ≤ 100, x, y ≥ 0

The minimum value of Z is 510 at D( 10,50)
Hence the supply from godown A are 10,50,40 quintals and from godown B are 50,0,0 quintals to shops D, E and F respectively and the minimum cost is ₹ 510

Question 7.
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500 L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:

Assuming that the transportation cost of 10 litres of oil is ₹ 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let x litre of oil be supplied from depot A to pump D and y litre of oil from depot B to pumb E. Then 7000 – (x + y) litre of oil will be transported to pump F.
The oil supplied from B to D, E and F are 4500 – x, 3000 – y, x + y – 3500

The transportation cost of 10 litres of oil for 1 km is ₹ 1.
∴ Transportation cost

Z = 0.3 x + 0.1 y + 3950
The transported quantity of oil is non negative.
i.e., x ≥ 0, y ≥ 0, 7000 – (x + y) ≥ 0
4500 – x ≥ 0, 3000 – y ≥0, x + y – 3500 ≥ 0
i.e., x, y ≥ 0
x + y ≤ 7000
x + y ≥ 3500
x ≤ 4500
y ≤ 3000
Hence the L.P.P. is
Minimise Z = 0.3x + 0. 1 y + 3950 subject to the constraints x ≤ 4500, y ≤ 3000, x + y ≤ 7000, x + y ≥ 3500, x, y ≥ 0

The feasible region is shaded in the figure. We use comer point method to find the minimum value of Z.

The minimum of Z is 4400 at E (500, 3000)
When x = 500 and y = 3000 500 l, 3000 l, 3500 l of oil should be trans¬ported from depot A to pumps D, E and F and 4000 l, 0 l, 0 l are transported from B to pumps D, E and F

Question 8.
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs atleast 240 kg of phosphoric acid, atleast 270 kg of potash and atmost 310 kg of chlorine.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Solution:
Let x bags of brand P and y bags of brand Q be used in the garden.
We have the data as

Phosphoric acid constraint: x + 2y ≥ 240
Potash constraint: 3x + 1.5y ≥ 270
Chlorine constraint: 1.5x + 2y ≤ 310
Quantity of nitrogen : Z = 3x + 3.5y
The L.P.P. is Minimise Z = 3x + 3.5y
subject to the constraints x + 2y ≥ 240, 3x + 1.5y ≥ 270, 1.5x + 2y ≤ 310, x, y ≥ 0.

The feasible region is shaded in the figure.

∴ Minimum value of Z is 470 at C(40, 100)
Hence 40 bags of brand P and 100 bags of brand Q are used.
The minimum amount of nitrogen is 470.

Question 9.
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs atleast 240 kg of phos-phoric acid, atleast 270 kg of potash and atmost 310 kg of chlorine. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Solution:
Let x bags of brand P and y bags of brand Q be used in the garden.
We have the data as

Phosphoric acid constraint: x + 2y ≥ 240
Potash constraint: 3x + 1.5y ≥ 270
Chlorine constraint: 1.5x + 2y ≤ 310
Quantity of nitrogen : Z = 3x + 3.5y
The L.P.P. is Minimise Z = 3x + 3.5y
subject to the constraints x + 2y ≥ 240, 3x + 1.5y ≥ 270, 1.5x + 2y ≤ 310, x, y ≥ 0.

The feasible region is shaded in the figure.

The maximum value of Z is 595 obtained at A(140,50)
Hence 140 bags of brand P and 50 bags of brand Q are used.
The maximum amount of nitrogen is 595.

Question 10.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is utmost half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by utmost 600 units. If the company makes profit of ₹ 12 and ₹ 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Solution:
Let the number of dolls of type A be x and that of type B be y.
We have the data as

Total dolls constraint: x + y ≤ 1200
Doll A constraint: x ≤ 3y + 600 or x – 3y ≤ 600
Profit function : Z = 12x +16y
The L.P.P is maximise Z = 12x + 16y
subject to the constraints
x + y ≤ 1200, x – 3y ≤ 600, y ≤ \(\frac { x }{ 2 }\), x ≥ 0, y ≥ 0

The maximum value of Z is 16000 at C(800,400) Hence 800 dolls of type A and 400 dolls of type B should be manufactured to get maximum profit.
The maximum profit is ₹ 16000.

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-2/

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2

Class 12 Maths Ex 13.2 NCERT Solutions  Question 1.
If P(A) = \(\frac { 3 }{ 5 }\) and P(B) = \(\frac { 1 }{ 5 }\), find P(A∩B) if A and B are independent events.
Solution:
A and B are independent if
P (A ∩ B) = P(A) P(B) = \(\frac { 3 }{ 5 } \times \frac { 1 }{ 5 } =\frac { 3 }{ 25 } \)

Ex 13.2 Class 12 NCERT Solutions Question 2.
Two cards are drawn at random and without replacement from a pack of 52 playing cards.Find the probability that both the cards are black.
Solution:
Let E : first card drawn is black
F : second card drawn is black
P(E ∩ F) = P(E) P(F|E)
P(E) = Probability of getting a black card \(\frac { 26 }{ 52 }\)
P(F|E) = Probability of getting a black card in the second draw given that the first card is black = \(\frac { 21 }{ 51 }\)
P(E ∩ F) = \(\frac { 26 }{ 52 }\) x \(\frac { 21 }{ 51 }\) = \(\frac { 25 }{ 102 }\)

Question 3.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution:
S = {12 good oranges, 3 bad oranges),
n(S) = 15
P (a box is approved) = \(\frac { C(12,3) }{ C(15,3) } =\frac { 12\times 11\times 10 }{ 15\times 14\times 13 } =\frac { 44 }{ 91 }\)

Question 4.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not
Solution:

Question 5.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the event, ‘the number is red’. Are A and B independent?
Solution:

Question 6.
Let E and F be the events with P(E) = \(\frac { 3 }{ 5 }\), P (F) = \(\frac { 3 }{ 10 }\) and P (E ∩ F) = \(\frac { 1 }{ 5 }\). Are E and F independent?
Solution:
P(E) = \(\frac { 3 }{ 5 }\), P(F) = \(\frac { 3 }{ 10 }\),
∴ P (E) x P (F) = \(\frac { 3 }{ 5 } \times \frac { 3 }{ 10 } =\frac { 9 }{ 50 }\)
P(E ∩ F) ≠ P(E) x P(F)
∴ The event A and B are not independent.

Question 7.
Given that the events A and B are such that P(A) = \(\frac { 1 }{ 2 }\), P(A∪B) = \(\frac { 3 }{ 5 }\) and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent.
Solution:

Question 8.
Let A and B independent events P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A∩B)
(ii) P(A∪B)
(iii) P (A | B)
(iv) P(B | A)
Solution:
P (A) = 0.3,
P (B) = 0.4
A and B are independent events
(i) ∴ P (A ∩ B) = P (A). P (B) = 0.3 x 0.4 = 0.12.

(ii) P(A∪B) = P(A) + P(B) – P(A).P(B)
= 0.3 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58.

(iii) P (A | B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.12}{0.4}=0.3\)

(iv) P(B | A) = \(\frac{P(A \cap B)}{P(A)}=\frac{0.12}{0.3}=0.4\)

Question 9.
If A and B are two events, such that P (A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\), and P(A ∩B) = \(\frac { 1 }{ 8 }\). Find P (not A and not B)
Solution:
P (A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\), and P(A ∩B) = \(\frac { 1 }{ 8 }\)
P(A) P(B) = \(\frac { 1 }{ 4 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 8 }\)
P(A).P(B) = P(A ∩ B)
∴ A and B are independent events
Since A and B are independent events, A’ and B’ are also independent events.
∴ P (A’ ∩ B’) = P (A’). P (B’)
= [1 – P(A)] [1 – P(B)]
= \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{2}\right)=\frac{3}{4} \times \frac{1}{2}=\frac{3}{8}\)

Question 10
Events A and B are such that P(A) = \(\frac { 1 }{ 2 }\), P(B) = \(\frac { 7 }{ 12 }\) and P (not A or not B) = \(\frac { 1 }{ 4 }\). State whether A and Bare independent
Solution:

Hence A and B are not independent events.

Question 11
Given two independent events A and B such that P (A) = 0.3, P(B) = 0.6. Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P (A or B)
(iv) P (neither A nor B)
Solution:
P(A) = 0.3, P(B) = 0.6
A and B are independent events.
i. P(A and B) = P(AB) = P(A).P(B)
= 0.3 x 0.6 = 0.18

ii. P(A and not B) = P(A B’ ) = P(A) P( B’)
= P(A) [1 – P(B)] = 0.3 (1 – 0.6)
= 0.3 x 0.4 = 0.12

iii P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(AB)
= 0.3 + 0.6 – 0.18 from (i)
= 0.72

iv. P(neither A nor B) = P( A’ B’)
= P(A’)P(B’)
= [1 – P(A)] [1 – P(B)]
= [ 1 – 0.3] [1 – 0.6]
= 0.7 x 0.4 = 0.28

Another Method:
P(neither A nor B) = P(A’ B’)
= P(A’ ∩ B’) = P((A ∪ B)’)
= 1 – P(A ∪ B) = 1 – 0.72 from (iii) = 0.28

Question 12
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
P(getting an odd number when a die is 1 tossed) = \(\frac { 1 }{ 2 }\)
P(getting an even number when a die is 1 tossed) = \(\frac { 1 }{ 2 }\)
Let A : getting even number in all the 3 throws
P(A) = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 8 }\)
P(getting an odd number atleast once) = P(not getting even number in the 3 throws)
= 1 – P(A) = 1 – \(\frac { 1 }{ 8 }\) = \(\frac { 7 }{ 8 }\)

Question 13
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
Solution:
Let R : red ball is drawn
B : black ball is drawn
P(R) = \(\frac { 8 }{ 18 }\), P(B) = \(\frac { 10 }{ 18 }\)
i. P(both are red) = P(1^st ball is red and 2^nd ball is red)
= P(RR) = P(R). P(R),
since the events are independent
= \(\frac { 8 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 6 }{ 81 }\)

ii. P(1^st ball is black and 2^nd ball is red)
= P(BR) = P(B) . P(R)
since the events are independent
\(\frac { 10 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 20 }{ 81 }\)

iii. P(one of them is black and the other is red) = P(1^st ball is black & 2^nd ball is red or 1^st ball is red & 2^nd ball is black) = P(BR or RB)
= P(BR) + P(RB)
= P(B) P(R) + P(R) P(B)
= 2 x \(\frac { 10 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 40 }{ 81 }\)

Question 14
Probability of solving specific problem independently by A and B are \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 3 }\) respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.
Solution:

Question 15
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace ’

(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’

(iii) E: ‘the card drawn is a king or queen ’
F: ‘the card drawn is a queen or jack ’.
Solution:
(i) E : card drawn is a spade
F : card drawn is an ace
E ∩ F : card drawn is an ace of spade

Hence E, F are independent events

(ii) E : card drawn is black
F : card drawn is a king
E ∩ F : card drawn is a black king

Hence E and F are independent events.

(iii) E : card drawn is king or queen
F : card drawn is queen or jack
E ∩ F : card drawn is queen

Hence E and F are not independent events.

Question 16
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspaper, find the probability that she reads English newspapers.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
Let H : student reads Hindi newspaper
E : student reads English newspaper

Question 17
The probability of obtaining an even prime number on each die when a pair of dice is rolled is
(a) 0
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 1 }{ 12 }\)
(d) \(\frac { 1 }{ 36 }\)
Solution:
(d) n(S) = 36
The sample space contains 36 simple events
Let E: getting even prime number on both dice
∴ E = {(2, 2)} ∴ P(E) = \(\frac { 1 }{ 36 }\)

Question 18
Two events A and B are said to be independent, if
(a) A and B are mutually exclusive
(b) P(A’B’) = [1 – P(A)] [1 – P(B)]
(c) P(A) = P(B)
(d) P (A) + P (B) = 1
Solution:
A and B are independent events if P(A ∩ B) = P(A) P(B)
P( A’B’) = P( A’ ∩ B’) = P((A ∪ B)’)
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – P(A) – P(B) + P(A) P(B)
= [1 – P(A)] – P(B)[1 – P(A)]
= [1 – P(A)][1 – P(B)]
= P(A’)P(B’)
If A and B are mutually exclusive, then A and B are not independent events.
If P(A) = P(B), then A and B need not be independent events.
If P(A) + P(B) = 1, then A and B are complements of each other. Then also A and B are not independent events.

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1/

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1

Question 1
Given that E and Fare events such that P (E) = 0.6, P (F) = 0.3 and P(E∩F) = 0.2 find P(E|F) and P (F|E).
Solution:
Given: P (E)=0.6, P (F)=0.3, P (E ∩ F)=0.2
P(E|F) = \(\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.3 } =\frac { 2 }{ 3 }\)
P(F|E) = \(\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.6 } =\frac { 1 }{ 3 }\)

Question 2
Compute P(A|B) if P(B)=0.5 and P (A∩B) = 0.32.
Solution:
Given: P(B) = 0.5, P(A∩B) = 0.32
P(A|B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.50 } =\frac { 32 }{ 50 }\) = 0.64

Question 3.
If P (A) = 0.8, P (B)=0.5 and P(B/A)=0.4, find
(i) P(A∩B)
(ii) P(A/B)
(iii)P(A∪B)
Solution:
(i) P(B/A) = \(\frac { P(A\cap B) }{ P(A) } \Rightarrow 0.4=\frac { P(A\cap B) }{ 0.8 }\)
∴ P(A∩B) = 0.4 x 0.8 = 0.32

(ii) P(A/B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.5 } =\frac { 32 }{ 50 } =\frac { 16 }{ 25 }\)

(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 1.30 – 0.32 = 0.98

Question 4.
Evaluate P(A∪B) if 2P(A) = P(B) = \(\frac { 5 }{ 13 }\) and P(A|B) = \(\frac { 2 }{ 5 }\).
Solution:

Question 5.
If P(A) = \(\frac { 6 }{ 11 }\),P(B) =\(\frac { 5 }{ 11 }\) and P(A∪B) = \(\frac { 7 }{ 11 }\),
Find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Solution:

Question 6.
A coin is tossed three times, where
(i) E: head on third toss F: heads on first two tosses.
(ii) E: at least two heads F : at most two heads
(iii) E: at most two tails F: at least one tail
Solution:
The sample space S = {HHH, HHT, THH, TTH, THH, THT, TTT}

Question 7.
Two coins are tossed once
(i) E: tail appears on one coin F: one coin shows head
(ii) E: no tail appears F: no head appears.
Solution:
The Sample space S = {HH, TH, HT, TT}

Question 8.
A die is thrown three times.
E: 4 appears on the third toss
F: 6 and 5 appears respectively on first two tosses.
Solution:

Question 9.
Mother, father and son line up at random for a family picture:
E: son on one end, F: father in middle
Solution:
Mother (m), Father (f) and son (s) line up at random E : son on one end: {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
F: Father in middle: {(m, f,s), (s, f, m), (s, f, m)}
E∩F = {(m, f, s), (s, f, m)}

Question 10.
A Mack and a red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution:
Let S be the sample space. Then
n(S) = 36
a. E : event of getting a sum greater than 9
E = {(4, 6), (5, 5), (5, 6), (6,4), (6, 5), (6, 6)}
F: event of getting 5 on black die.
F = {(5, 1), (5, 2), (5, 3), (5,4), (5, 5), (5, 6))
E ∩ F = {(5, 5), (5, 6))
\(\mathrm{P}(\mathrm{F})=\frac{6}{36}, \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36}\)
\(P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\left(\frac{2}{36}\right)}{\left(\frac{6}{36}\right)}=\frac{2}{6}=\frac{1}{3}\)

b. E : event of getting a sum 8
E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F : event of getting a number less than 4 on red die
F= {(1, 1),(2, 1),(3, 1),(4, 1),(5, 1),(6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1,3), (2,3), (3, 3), (4, 3), (5, 3), (6, 3)}
E ∩ F = {(5,3),(6,2)}
\(P(E \cap F)=\frac{2}{36}, P(F)=\frac{18}{36}\)
∴ P(E|F) = \(\frac{P(E ∩ F)}{P(F)}\) = \(\frac{(\frac{2}{36})}{(\frac{18}{36})}\) = \(\frac{1}{9}\)

Question 11
A fair die is rolled. Consider events E = {1,3,5} F = {2,3} and G = {2,3,4,5}, Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F) |G) and P (E ∩ F)|G)
Solution:

Question 12.
Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl,
(ii) at least one is girl?
Solution:
Let B : The elder boy
G : The elder girl
b : The younger boy
g : The younger girl
The sample space S = {(B, h), (B, g), (G, b),(G, g)}

i. Let E : both children are girls
F : the youngest is a girl

ii. Let E : both children are girls
F : atleast on child is a girl

Question 13.
An instructor has a question bank consisting of 300 easy True/ False questions, 200 difficult True/ False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution:

Question 14.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution:
Let S be the sample space.
∴ n(S) =36
Let E : getting different numbers on the dice
∴ n(E) = 30 ∴P(E) = \(\frac { 30 }{ 36 }\)
Let F : getting the sum of numbers on the dice is 4
∴ F= {(1, 3), (2, 2), (3, 1))
n(F) = 3 ∴ P(F) = \(\frac { 3 }{ 36 }\)
E ∩ F = {(1,3),(3, 1)}
n(E ∩ F) = 2 ∴ P(E ∩ F) = \(\frac{2}{36}\)
\(\mathrm{P}(\mathrm{F} \mid \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})}=\frac{\left(\frac{2}{36}\right)}{\left(\frac{30}{36}\right)}=\frac{1}{15}\)

Question 15.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’ given that ‘at least one die shows a 3’.
Solution:
The sample space S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, H). (1, T), (2, H). (2, T), (4, H). (4, T), (5, H), (5, T), (6, 1),(6, 2),(6, 3), (6,4),(6, 5),(6, 6))
Let E: coin shows tail
F : atleast one die shows 3
Now E ∩ F = Φ or P(E ∩ F) = 0
P(E|F) = 0 since P(E ∩ F) = 0

Question 16.
If P(A) = \(\frac { 1 }{ 2 }\), P (B) = 0 then P (A | B) is
(a) 0
(b) \(\frac { 1 }{ 2 }\)
(c) not defined
(d) 1
Solution:
P(A) = P(B) = 0
∴ P(A∩B) = 0
∴ P(A|B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0 }{ 0 }\)
Thus option C is correct.

Question 17.
If A and B are events such that P(A | B) = P(B | A) then
(a) A⊂B but A≠B
(b) A = B
(c) A∩B = φ
(d) P(A) = P(B)
Solution:
P(A | B) = P(B | A)
Thus option (d) is correct.
\(\frac { P(A\cap B) }{ P(B) } =\frac { P(B\cap A) }{ P(A) }\)
⇒ P(A) = P(B)

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-5/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.5

Question 1.
Find [a, b, c] if \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}-2 \hat{k}\)
Solution:
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ 2 & -3 & 1 \\ 3 & 1 & -2 \end{array}\right|\) = 1(6 – 1)+2(- 4 – 3) + 3(2 + 9) = 1(5) + 2(- 7) + 3(11) = 24.

Question 2.
Show that the vectors \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}=-2 \hat{i}+3 \hat{j}-4\hat{k}\) and \(\vec{c}=\hat{i}-3\hat{j}+5 \hat{k}\) are coplanar.
Solution:
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|\)
= 1(15 – 12)+2(- 10 + 4) + 3(6 – 3)
= 1(3) + 2(- 6) + 3(3) = 0.

Question 3.
Find λ, if the vectors \(\hat{i}-\hat{j}+\hat{k}, 3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\hat{i}+\lambda \hat{j}-3 \hat{k}\)are coplanar.
Solution:
Let \(\vec{a}=\hat{i}-\hat{j}+\hat{k}\), \(\vec{b}=3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{c}=\hat{i}+\lambda \hat{j}-3 \hat{k}\) be the given vectors.
Since the given vectors are coplanar \([\vec{a}, \vec{b}, \vec{c}]\) = 0
i.e., \(\left|\begin{array}{ccc} 1 & -1 & 1 \\ 3 & 1 & 2 \\ 1 & \lambda & -3 \end{array}\right|\) = 0
1(- 3 – 2λ) + 1(- 9 – 2) + 1(3λ – 1) = 0
– 3 – 2λ – 11 + 3λ – 1 = 0 ⇒ λ = 15

Question 4.
i. If c₁ = 1 and c₂ = 2, find c₃ which makes \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.
ii. Then if c₂ = – 1 and c₃ = 1, show that no value of c₁ can make \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.
Solution:
i. Given that \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.
∴ \([\vec{a}, \vec{b}, \vec{c}]\) = 0

Question 5.
Consider the 4 points A, B, C and D with position vectors \(4 \hat{i}+8 \hat{j}+12 \hat{k}, 2 \hat{i}+4 \hat{j}+6 \hat{k}\) \(3 \hat{i}+5 \hat{j}+4 \hat{k} \text { and } 5 \hat{i}+8 \hat{j}+5 \hat{k}\).
i. Find \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}\).
ii. Show that the 4 points are coplanar.
Solution:

Question 6.
Find x such that the four points A(3, 2, 1), B(4, x, 5), C(4, 2, – 2) and D(6, 5, – 1) are coplanar
Solution:

ii. Since the 4 points A, B, C and D are coplanar \([\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}]\) = 0
\(\left|\begin{array}{ccc}
1 & (x-2) & 4 \\
1 & 0 & -3 \\
3 & 3 & -2
\end{array}\right|\) = 0
1(0 + 9) – (x – 2)(- 2+9) + 4(3 – 0) = 0
9 – 7(x – 2) + 12 = 0
9 – 7x + 14 + 12 = 0
⇒ 7x = 35
∴ x = 5

Question 7.
Show that the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar if \(\vec{a}\) + \(\vec{b}\), \(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\) are coplanar.
Solution: