Class 11 Maths Ncert Solutions Ex 16.3

Class 11 Maths Ncert Solutions Chapter 16 Ex 16.3

Question 1

Judge the outcome of the following assignment is valid or not.

S = {ω7,ω6,ω5,ω4,ω3,ω2,ω1}

 

Assignment ω1 ω2 ω3 ω4 ω5 ω6 ω7
(a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6
(b) 17 17 17 17 17 17 17
(c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7
(d) -0.1 0.2 0.3 0.4 -0.2 0.1 0.3
(e) 114 214 314 414 514 614 1514

 

Sol:

(a)

ω1 ω2 ω3 ω4 ω5 ω6 ω7
0.1 0.01 0.05 0.03 0.01 0.2 0.6

 

Each of the numbers such as p(ω1),p(ω2)..p(ω7) are positive in nature and are less than 1.

= p(ω7)+p(ω6)+p(ω5)+p(ω4)+p(ω3)+p(ω2)+p(ω1)

= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

= 1

The assignment is valid.

 

(b)

ω1 ω2 ω3 ω4 ω5 ω6 ω7
17 17 17 17 17 17 17

 

Each of the numbers such as p(ω1),p(ω2)..p(ω7) are positive in nature and are less than 1.

= p(ω7)+p(ω6)+p(ω5)+p(ω4)+p(ω3)+p(ω2)+p(ω1)

= 17 + 17 + 17 + 17 + 17 + 17 + 17

= 7 (17)

= 1

The assignment is valid.

 

(c)

ω1 ω2 ω3 ω4 ω5 ω6 ω7
0.1 0.2 0.3 0.4 0.5 0.6 0.7

 

Each of the numbers such as p(ω1),p(ω2)..p(ω7) are positive in nature and are less than 1.

= p(ω7)+p(ω6)+p(ω5)+p(ω4)+p(ω3)+p(ω2)+p(ω1)

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7

= 2.8

≠ 1

The assignment is not valid.

 

(d)

ω1 ω2 ω3 ω4 ω5 ω6 ω7
-0.1 0.2 0.3 0.4 -0.2 0.1 0.3

 

From the above table it is evident that p(ω1) & p(ω5) are negative.

Therefore,

The assignment is not valid.

 

(e)

ω1     ω2      ω3         ω4           ω5             ω6         &nbsnbsp;  ω7
114 214 314 414 514 614 714

p(ω7) = 1514 ˃ 1

The assignment is not valid.

 

 

Question 2

An experiment consists of tossing up of a coin, determine the probability of getting at least one tail?

Sol:

Sample space of the experiment when the coin is tossed twice.

S = {TT , TH , HT , HH }

Let the event P be the event of occurring at least 1 tail.

According to the question = { TT , TH , HT }

Therefore, P (P) = No.offavourableoutcomesTotalno.ofoutcomes

= n(P)n(S)

= 34

 

 

Question 3

An experiment consists of throwing of a dice in which the following events occurs:

(a) A prime number

(b) Number larger than 3

(c) Number less than or equal to 1

(d) Number more than 6

(e) Number less than 6

Sol:

Sample space: { 6 , 5 , 4 , 3 , 2 , 1 }

(a) Let us consider event P be the event of occurrence of prime number

P = { 5 , 3 , 2 }

=No.offavourableoutcomesTotalno.ofoutcomes

= n(P)n(S)

= 36

= 12

(b) Let us consider event Q be the event of occurrence of a number greater than or equals to 3

Q = { 6,5,4,3}

=No.offavourableoutcomesTotalno.ofoutcomes

= n(Q)n(S)

= 46

= 23

 

(c) Let us consider event R be the event of occurrence of a number less than or equals to 1

R = { 1 }

=No.offavourableoutcomesTotalno.ofoutcomes

= n(R)n(S)

= 16

 

(d) Let us consider event S be the event of occurrence of a number greater than 6

S = ϕ

=No.offavourableoutcomesTotalno.ofoutcomes

= n(S)n(S)

= 06

= 0

 

(e) Let us consider event T be the event of occurrence of a number less than 6

T = { 5 , 4 , 3 , 2 , 1 }

=No.offavourableoutcomesTotalno.ofoutcomes

= n(T)n(S)

= 56

 

Question 4

An experiment consists of a pack of 52 cards.

(a) Find the number of elements present in the sample space.

(b) Find the probability of ace of spades

(c) Find the probability of (i) an ace (ii) black card

Sol:

(a)  When a card is drawn from a deck of cards having a total of 52 cards , the total number of possible outcomes is 52 that i.e 13 cards of each kind ( clubs , spades , diamonds and hearts )

 

(b) Let us consider P be the event of drawing an ace card .

Number of favourable outcomes = n (P) = 4

P (E) =No.offavourableoutcomesTotalno.ofoutcomes

= n(P)n(S)

= 152

= 152

 

(c) Let us consider Q be the event of drawing an ace card.

Number of favourable outcomes = n (Q) = 4

P (E) =No.offavourableoutcomesTotalno.ofoutcomes

= n(Q)n(S)

= 452

= 113

 

Let us consider R be the event of drawing a black card.

Number of favourable outcomes = n (R) = 26

P (E) =No.offavourableoutcomesTotalno.ofoutcomes

= n(R)n(S)

= 2652

= 12

 

 

Question 5

An unbiased coin is marked 1 on one face while 6 on the other face , while the die is having markings such as 1,2,3,4,5,6 on its 6 faces .

Determine the probability that the sum of the numbers turning up is

(i) 3

(ii) 12

Sol:

According to the question, an unbiased coin is marked 1 on one of the face and 6 on the other face . Also, a die is having 6 faces with marking 1 to 6 respectively.

Sample space (S) = {(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (6,1) , (6,2) , (6,3) , (6,4) ,(6,5) ,(6,6)}

Therefore = n(s) = 12

(i) Let us consider P be the event of having the sum of the numbers turning up is 3 = P = { (1,2) }

Accordingly,  P (E) =No.offavourableoutcomesTotalno.ofoutcomes

= n(P)n(S)

= 112

 

(ii) Let us consider Q be the event of having the sum of the numbers turning up is 12 = P = { (6,6) }

Accordingly,  P (E) =No.offavourableoutcomesTotalno.ofoutcomes

= n(Q)n(S)

= 112

 

 

Question 6

A city council consists of 4 men and 6 women .Find the probability of selecting a woman among the council members if the selection is on random basis.

Sol:

According to the question, the city council consists of 4 men and 6 women . Council members are selected on random basis one at a time .

Sample space = 6+4 = 10

Let p be the event of selecting a women among the council members of a city council = n (P) = 6

P (E) = =No.offavourableoutcomesTotalno.ofoutcomes

= n(P)n(S)

= 610

= 35

 

 

Question 7

An experiment consists of tossing up of an unbiased coin 4 times. Each time head appears a person is awarded with Re.1 and each time a tail a turns up that person looses Rs.1.50. Determine sample space of different money of 4 tosses and the propability of each of the head and tail.

Sol:

According to the question , the coin is tossed 4 times. Each time head appears the person is awarded with Re.1 but if tails appears that person looses Rs.1.50.

At most 4 heads can turn up = Re. 1 + Re. 1 + Re. 1 + Re. 1

= Rs. 4 (Gain )

When 3 heads and 1 tail appears = Re. 1 + Re. 1 + Re. 1 – Rs.1.50

= Rs. 1.50

When 2 heads and 2 tail appears = Re. 1 + Re. 1 – Rs. 1.50 – Rs.1.50

= -Re. 1.00

When 1 heads and 3 tail appears = Re. 1 – Rs. 1.50 + Rs.1.50 – Rs.1.50

= -Rs. 3.50

At least no head appears , all the 4 times tail appears =  – Rs.1.50 – Rs.1.50 – Rs.1.50 – Rs.1.50

= Rs. – 6 (loss )

In the sample space total number of elements = 24 elements

= 16 elements

Sample space (S) = { TTTT ,  TTTH , TTHT , THTT , HTTT , THHT , THTH , HTHT , TTHH , HTTH , HHTT , THHH , HTHH , HHTH , HHHT , HHHH }

The person gains Rs. 4 for 4 head turn ups for the sample event { HHHH }

Probability ( Gain of Rs. 4 )  = 116

The person gains Rs. 1.50 for 3 head turn ups for the sample event { HHHT , HHTH , HTHH , THHH }

Probability ( Gain of Rs. 1.50 )  = 416 =      14

The person looses Re. 1 for 2 head turn ups for the sample event { HHTT , HTTH , TTHH , HTHT , THTH , THHT }

Probability ( Loss of Re. 1 )  = 616 =      38

The person looses Rs. 3.50 for 1 head and 3 tails turn ups  for the sample event { TTTT }

Probability ( Loss of Rs. 6 )  = 116

 

Question 8

An experiment consists of tossing up of 3 coins .Find out the probability of the following events:

(a) Three heads

(b) Two heads

(c) At least two heads

(d) At most two heads

(e) No heads

(f) Three tails

(g) Exactly two tails

(h) No tails

(i) At most two tails

Sol:

According to the question , an experiment consists of tossing up of three coins .

Sample space of total number of possible events = { HHH , HHT , HTH , THH , HTT , THT , TTH , TTT }

So, the total number of possible outcomes = n(s) = 8

P (A) = =No.offavourableoutcomesTotalno.ofoutcomes

(a) Let Q be the event of occurring three heads = { HHH }

Therefore , P(Q) = n(Q)n(S)=18

 

(b) Let R be the event of occurring two heads = { HHT , HTH ,THH }

Therefore , P(R) = n(R)n(S)=318

 

(c) Let S’ be the event of occurring at least 2 heads = { HHH ,HHT ,HTH ,THH }

Therefore , P(S’) = n(S)n(S)=48=12

 

(d) Let T be the event of occurring at most 2 heads = { HHT , HTH ,THH , HTT , THT , TTH , TTT }

Therefore , P(T) = n(T)n(S)=78

 

(e) Let U be the event of occurring no heads = { TTT }

Therefore , P(U) = n(U)n(S)=18

 

(f) Let V be the event of occurring 3 tails = { TTT }

Therefore , P(V) = n(V)n(S)=18

 

(g) Let W be the event of occurring exactly two tails = { HTT , THT , TTH }

Therefore , P(W) = n(W)n(S)=38

 

(h) Let X be the event of occurring no tails = { HHH }

Therefore , P(X) = n(X)n(S)=18

 

(i) Let Y be the event of occurring at most 2 tails = { HHH , HHT , HTH , THH , HTT , THT , TTH }

Therefore , P(Y) = n(Y)n(S)=78

 

 

Question 9

The probability of an event is 211 . What is the probability of getting not the above mentioned event.

Sol:

Let the event be P and P’

Probability ( P) = 211

Probability (P’ ) = 1 – 211 = 911

 

 

Question 10

From the word ‘ASSASSINATION’ a letter is chosen at a random basis. Then, determine the probability of getting (a) a vowel (b) an consonant

Sol:

Total number of letters in the word ASSASSINATION is 13

Therefore, n(S) = 13

(a) In the above word there are 6 vowels.

Probability (Vowel ) =  613

(b) In the above word there are 7 consonants .

Probability ( Consosnant ) = 713

 

 

Question 11

A person chooses 6 different natural numbers at random basis from numbers ranging from 1 to 20. If and only if this 6-digit number  matches with the number decided by the lottery committee , the person wins the prize. Determine the probability of wining the prize ?

Sol:

According to the question , total number of ways 6 numbers can be chosen from numbers ranging from 1 to 20

= 206C

= 20×19×18×17×16×151×2×3×4×5×6

= 38760

Therefore the total possibilities are 38760

Out of this 38760 there is only 1 chance to win the  lottery,

Probability = 138760

 

 

Question 12

Check whether the following probabilities are defined or not :

(i) P(Q) = 0.5 , P(R) = 0.7 , P(QR) = 0.6

(ii) P(Q) = 0.5 , P(R) = 0.4 , P(QR) = 0.8

Sol:

(i) P(Q) = 0.5 , P(R) = 0.7 , P(AB) = 0.6

E and F are two events such that P(E) ≤ P(F)

According to the question ,

P(QR) ˃ P(Q)

P(Q) and P(R) are not defined.

 

(ii) P(Q) = 0.5 , P(R) = 0.4 , P(QR) = 0.8

E and F are two events such that P(E) ≤ P(F)

According to the question ,

P(QR) ˃ P(Q)

P(Q) and P(R) are defined.

 

 

Question 13

  P(Q) P(R) P(QR) P(QR)
(i) 13 15 115 ….
(ii) 0.35      …………… 0.25 0.6
(iii) 0.5 0.35    ……… 0.7

Sol:

(i) P(QR)=P(Q)+P(R)P(QR)

= P(QR)=13+15115

= 715

(ii) P(QR)=P(Q)+P(R)P(QR)

= 0.6 = 0.35 + P(R) – 0.25

= P(R) = 0.6 – 0.35 +0.25

= P(R) = 0.5

(iii) P(QR)=P(Q)+P(R)P(QR)

= P(QR)=0.5+0.350.7

= P(QR)=0.15

 

 

Question 14

P(Q) = 35 and P(R) =  15. find the value of P( Q or R ) if Q and R are mutually exclusive events.

Sol:

According to the question Q and R are mutually exclusive events .

P(Q or R) = P(Q) + P(R)

= P( Q or R ) = 35 + 35

= 45

 

 

Question 15

If Q and R are events such that P(Q) = 14 and P(R) =  12 and P( Q and R) =18   . Find the value of P( Q and R )

(a) P( Q or R)  

(b) P( not Q and not R)

Sol:

(a) P(Q or R) = P(Q) + P(R) – P( Q and R)

= P(Q or R) = 14+1218 = 58

(b) P( Q or R) = P(QR) = 18

=    P( not Q and not R)  = P(QR)’

P( not Q and not R) = 1 – P(Q and R)

=    P( not Q and not R)  = 1 –  P(QR)

= 1 –  58

38

 

 

Question 16

Events Q and R are such that P(not Q or not R ) = 0.25 . Find out whether Q and R are mutually exclusive or not ?

Sol:

Given, P ( not Q or not R ) = 0.25

= P(QR) = 0.25

= P(QR) = 0.25

= P(QR) = 1- P(QR)’

=   P(QR) = 1- 0.25

=     P(QR) = 0.75

P(QR) ≠ ϕ

 

 

Question 17

Events Q and R are such that P(Q) = 0.42 , P® = 0.48 & P(Q and R) = 0.16. Find:

(a) P( not Q)

(b) P(not R)

(c) P( Q or R)

Sol:

(a) P(not Q) = 1 – P(Q) = 1 – 0.42 = 0.58

(b) P(not R) = 1 – P(R) = 1 – 0.42 = 0.52

(c) P( Q or R) = P(Q) + P(R) – P( Q and R)

= 0.42 + 0.48 – 0.16 = 0.74

 

 

Question 18

In class XII of a school , 40% of the students studies mathematics & 30% of the students studies biology. 10% of the students of the class studies both maths and biology. If a student is selected as a random from the class , determine the probability of the student studying mathematics or biology.

Sol:

Let Q be the event in which students studies mathematics and R be the event in which student’s studies biology.

Now,

P(Q) = 40% = 40100 = 25

P(R) = 30% = 30100 = 310

P(Q and B) = 10100 = 110

We know that P( Q or R)  = P(Q)  + P(R) – P( Q and R)

= P( Q or R) = 25 + 310110

= 0.6

Therefore , the probability that the students studying mathematics or biology is 0.6

 

 

Question 19

In an entrance test that is based on the basis of two examinations, the probability of a random student passing the examination is 0.8 and the probability of the candidate passing the second examination is 0.7. The probability of passing at least one of the examination is 0.95. find out the probability of passing both  the examinations?

Sol:

Let Q and R be the events of candidates passing the first and second examinations respectively.

Accordingly to the question ,

P( Q or R ) = P(Q) + P(R) – P( Q and R)

= 0.95 = 0.8 + 0.7 – P( Q and R)

= P( Q and R)  = 1.5 – 0.95

= P( Q and R)  = 0.55

Hence , the probability of the candidates passing both the examinations is 0.55

 

 

Question 20

The probability that a student will pass the final examination in oth English and Hindi is 0.5. The probability of passing the neither of the subjects is 0.1. If the probability of passing the subject English alone is 0.75 then, find out the probability of passing in the Hindi subject?

Sol:

Let Q and R be the events of passing English and Hindi examinations respectively .

P( Q and R  ) =0.5

P( not Q and not R) = 0.1

P(Q) = 0.75

Therefore ,

=  ( QR)’ = (QR)

= P(QR)’ = P(QR)

= P(QR) = 1 – (QR)’ = 1 – 0.1 = 0.9

We know,

P( Q or R) = P( Q) + P(R) – P( Q and R)

= 0.9 = 0.75 + P(B) – 0.5

= P(B) = 0.65

Therefore the probability of passing the Hindi subject alone in the final examination is 0.65

 

 

Question 21

A class of strength 60 students , 30 students opted for NCC, 32 opted for NSS and 24 students opted for both NCC and NSS. Find the probability if one student is selected at a random that:

(a) Candidate opted out for NCC or NSS

(b) Candidates opted neither NCC nor NSS

(c) Candidates opted for NSS but not NCC

 Sol:

Let Q be the event in which students opted for NCC whereas R be the event in which students in which students opted for NSS

Total strength of the class = 60

Students opted for NCC alone = 30

P( Q) = 3060=12

Students opted for NSS = 32

P(R) = 3260=815

Number of students opted for both NCC and NSS = 24

P( Q and R) = 2460=25

(a) P(Q and R) = P(Q) + P(R) – P( Q and R)

=  P(Q and R) = 12 + 81525

= P(Q and R) = 1930

Therefore , the students opted for both NCC and NSS is  1930

(b) P(not Q and not R) = P(A’ and B’)

= P(AB)

= P(AB)’

= 1 – P(AB)’

= 1 – P(Q or R)

= 1 – 1930

= 1130

Probability that the students opted neither for NCC nor NSS is 1930

(c) The above mentioned problem can be best visualized with the help of vein diagram.

Untitled

It is clear from the above diagram that the number of students who opted for NCC but not NSS

Thus,

= n(R-Q) = n( R) – n(AB)

= 32-24

= 8

Therefore the probability of the students who have opted for NCC but not NSS is

860 = 215