NCERT Grade 12-Probability-Answers

NCERT Solutions for Class 12 Maths

Find 100% accurate solutions for NCERT Class XII Maths. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

Sign Up to Download as PDF

1.    Given: P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2

2.    Given: P (B) = 0.5 and P (A ∩ B) = 0.32

3.    (i) P (A ∩ B) = P(B|A) . P (A) = 0.4 x 0.8 = 0.32

(ii) 

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

4.    

5.    (i) P (A ∪ B) = P(A) + P (B) – P (A ∩ B)

6.    A coin tossed three times, i.e.,

S = (TTT, HTT, THT, TTH, HHT, HTH, THH, HHH)

⇒ n(S) = 8

(i) E : heads on third toss

E = (TTH, HTH, THH, HHH)

⇒ n(E) = 4

F : heads on first two tosses

F = (HHT, HHH)

⇒ n(F) = 2

∴ E ∩ F = (HHH)

⇒ n(E ∩ F) = 1

(ii) E : at least two heads

E = (HHT, HTH, THH, HHH)

⇒ n(E) = 4

F : at most two heads

F = (TTT, HTT, THT, TTH, HHT, HTH, THH)

⇒ n(F) = 7

(iii) E : at most two tails

E = (HTT, THT, TTH, HHT, HTH, THH, HHH)

⇒ n(F) = 7

F : at least one tail

F = (TTT, HTT, THT, TTH, HHT, HTH, THH)

⇒ n(F) = 7

∴ E F = (HTT, THT, TTH, HHT, HTH, THH)

7.    S = (HH, TH, HT, TT) ⇒ n(S) = 4

(i) E : tail appears on one coin

E = (TH, HT) ⇒ n(E) = 2

(ii)  E : no tail appears

8.    Since a dice has six faces. Therefore n(S) = 6 x 6 x 6 = 216

E = (1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) x (4)

F = (6) x (5) x (1, 2, 3, 4, 5, 6)

⇒ n(F) = 1 x 1 x 6 = 6

9.    S = (MFS, MSF, SFM, SMF, FMS, FSM) ⇒ n(S) = 6

E : Son on one end

E = (MFS, SFM, SMF, FMS) ⇒ n(E) = 4

F : Father in middle

F = (MFS, SFM) ⇒ n(F) = 2

10.  (a) n(S) = 6 x 6 = 36

Let A represents obtaining a sum greater than 9 and B represents black die resulted in a 5.

A = (46, 64, 55, 36, 63, 45, 54, 65, 56, 66) ⇒ n(A) = 10

(b) Let A denoted the sum is 8

∴ A = {(2, 6). (3, 5), (4, 4), (5, 3), (6, 2)}

B = Red die results in a number less than 4, either first or second die is red

∴ B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}

11.  S = (1, 2, 3, 4, 5, 6) ⇒ n(S) = 6

E = (1, 3, 5) F = (2, 3) G = (2, 3, 4, 5)

12.  Let first and second girl are denoted by G1 and G2 and boys B1 and B2.

∴ S = {(G1G2), (G1B2), (G2B1), (B1B2)}

Let A = Both the children are girls = (G1G2)

B = Youngest child is girl = {(G1G2), (B1G2)}

C = at least one is a girl = {(G1B2), (G1G2), (B1G2)}

13.  Number of easy True/False questions = 300

Number of difficult True/False questions = 200

Number of easy multiple choice questions = 500

Number of difficult multiple choice questions = 400

Total number of all such questions = = 1400

Let E represents an easy question and F represents a multiple choice question.

14.  S = (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)

(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)

∴ n(S) = 36

Let A represents the event “the sum of numbers on the dice is 4” and B represents the event “the two numbers appearing on throwing two dice are different”.

Therefore, A = {(1, 3), (2, 2), (3, 1)} ⇒ n(A) = 3

15.  S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

(1, H), (2, H), (3, H), (4, H), (5, H), (1, T), (2, T), (3, T), (4, T), (5, T)}

∴ n(S) = 20

P (first die shows a multiple of 3)

P (first die shows a number which is not a multiple of 3) 

Let A = the coin shows a tail = {(1, T), (2, T), (4, T), (5, T)}

B = at least one die shows a 3 = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}

16.  

Therefore, option (C) is correct.

17.  

Therefore, option (D) is correct.

18.  As A and B are independent events.

Therefore, P(A ∩ B) = P (A). P (B)

19.  S = 52 cards ⇒ n(S) = 52

Two cards are drawn without replacement.

A = {26 black cards} ⇒(A) = 26

20.  S = {12 good oranges, 3 bad oranges}

⇒ n(S) = 15

Probability that first orange drawn is good 

Probability that second orange is drawn is good 

Probability that third orange is drawn is good when both the first and second are good 

P (a box is approved) 

21.  S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

⇒ n(S) = 12

Head appears on the coin = A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)} ⇒ n(A) = 6

22.  

23.  

Therefore, E and F are not independent events.

24.  

25.  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) = 0.03 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58

26.  

Thus, A and B are independent events.

Therefore, ‘not A’ and ‘not B’ are independent events.

Hence, P (not A and not B) = P (not A). P (not B)

27.  

28.  P (A) = 0.3, P (B) = 0.6

A and B are independent events.

(i) P (A and B) = P (A). P (B) = 0.3 x 0.6 = 0.18

(ii) P (A and not B) =  = P (A) – P(A ∩ B) = 0.3 – 0.18 = 0.12

(iii) P (A or B) = P (A) + P (B) – P (A and B) = 0.3 + 0.6 – 0.18 = 0.9 – 0.18 = 0.72

(iv) P (neither A nor B) = P [not (A ∪ B)] = 1 – P (A ∪ B) = 1 – 0.72 = 0.28.

29.  S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6

Let A represents an odd number.

Now P (at least one success) = 1 – P (X = 0)

30.  S = (10 black balls, 8 red balls) ⇒ n(S) = 18

Let drawing of a red ball be a success.

(i) P (both are red ball) = P (A). P (B) =

(ii) P (first is black ball and second is red) 

(iii) P (one of them is black and other is red) 

31.  

32.  S = {All the 52 cards} ⇒ n(S) = 52

(i) E = {13 spades} ⇒ n(E) = 13

Hence, E and F are independent events.

(ii) E = {26 black cards} ⇒ n(E) = 26

Hence, E and F are independent events.

(iii) E = {4 kings, 4 queens} ⇒ n(E) = 8

Hence, E and F are not independent events.

33.  Let A represents the student reading Hindi newspaper and B represents the students reading English newspaper.

(a) P (neither Hindi nor English newspaper is read) = 1 – P (A or B)

34.  n(S) = 36

Let A represents an even prime number on each dice.

Hence option (D) is correct.

35.  P (A’ and B’)

= [1 – P (A)] . [1 – P (B)]

= P (A’). P (B’)

Hence, option (B) is correct

36.  Case (i) : S1 = {5 red balls, 5 black balls}

⇒ n(S1) = 10

Let us draw a red balls first, i.e., A1 = {5 red balls}

⇒ n(A1) = 5

Now after adding 2 balls of the same colour, i.e., when the first draw gives a red ball, two additional red balls are put in the urn so that its contents are 7 (5 + 2) red and 5 black balls. When the first draw gives a black ball, two additional black balls are put in the urn so that its contents are 5 red and 7 (5 + 2) black balls.

Total balls = S2 = {7 red balls, 5 black balls}

⇒ n(S2) = 12

Let us draw a red balls first, i.e., A1 = {7 red balls}

⇒ n(A2) = 7

∴ P (a red ball is drawn) 

Case (ii) : When a black ball is drawn, i.e., A2 = {5 red balls}

Now after adding 2 balls of the same colour, i.e.,

S2 = {5 red balls, 7 black balls}

⇒ n(S2) = 12

Let us draw a red balls first, i.e., A2 = {5 red balls}

∵ P (a red ball is drawn)

Therefore, required probability in both cases 

37.  Let A be the event hat ball drawn is red and let E1 and E2 be the events that the ball drawn is from the first bag and second bag respectively.

38.  Let E1 = the examinee knows the answer, E2 = the examinee guesses the answer and

39.  Let E1 = students residing in the hostel, E2 = day scholars (not residing in the hostel) and

A = the examinee answers correctly

Since E1 and E2 are mutually exclusive events and exhaustive events, and if E2 has already occurred, then the examinee guesses, therefore the probability that he answers correctly given that he has made a guess is i.e., 

And P(A|E1) = P (answers correctly given that he knew the answer) = 1

Therefore, by Bayes’ theorem,

40.  Let E1 = The person selected is suffering from certain disease, E2 = The person selected is not suffering from certain disease and A = The doctor diagnoses correctly

41.  Let E1 = a two headed coin, E2 = a biased coin, E3 = an unbiased coin and A = A head is shown

42.  Let E1 = Person chosen is a scooter driver, E2 = Person chosen is a car driver, E3 = Person chosen is a truck driver and A = Person meets with an accident

Since there are 12000 persons, therefore,

It is given that P(A|E1) = P (a person meets with an accident, he is a scooter driver) = 0.01

Similarly, P(A|E2) = 0.03 and P(A|E3) = 0.15

To find: P (person meets with an accident that he was a scooter driver)

Therefore, by Bayes’ theorem,

43.  

44.  Given: P (G1) = 0.6, P (G2) = 0.4

Let P denotes the launching of new product.

45.  Let E1 = 5 or 6 appears on a die, E2 = 1, 2, 3 or 4 appears on a die and A = A head appears on the coin.

Now 

Now P(A|E1) Probability of getting a head on tossing a coin three times,

when E1 has already occurs = P (HTT) or P (THT) or P (TTH)

when E2 has already occurred = 

∴ P (there is exactly one head given that 1, 2, 3 or 4 appears on a die)

46.  Let E1 = the item is manufactured by the operator A, E2 = the item is manufactured by the operator B, E3 = the item is manufactured by the operator C and A = the item is defective

Now 

Now P(A|E1) = P (item drawn is manufactured by operator A) = 

Similarly, 

Now Required probability = Probability that the item is manufactured by operator A given that the item drawn is defective

47.  Let E1 = the missing card is a diamond, E2 = the missing card is a spade, E3 = the missing card is a club, E4 = the missing card is a heart and A = drawing of two heart cards from the remaining cards.

Now 

P= P (drawing 2 heart cards given that one diamond card is missing) = 

Similarly, 

By Bayes’ theorem,

48.  Let A be the event that the man reports that head occurs in tossing a coin and let E1 be the event that head occurs and E2 be the event head does not occur.

P(A|E1) = P (A reports that head occurs when head had actually occur red on the coin) = 

P(A|E2) = P (A reports that head occurs when head had not occur red on the coin) = 

By Bayes’ theorem,

Hence, option (A) is correct.

49.  

Hence, option (C) is correct.

50.  (i) P (0) + P (1) + P (2) = 0.4 + 0.4 + 0.2 = 1

Therefore, it is a probability distribution.

(ii) P (3) = –0.1 which is not possible.

Therefore, it is not a probability distribution.

(iii) P (–1) + P (0) + P (1) = 0.6 + 0.1 + 0.2 = 0.9 1

Therefore, it is not a probability distribution.

(iv) P (3) + P (2) + P (1) + P (0) + P (–1) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 1

Therefore, it is not a probability distribution.

51.  There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

52.  Let h denotes the number of heads and denotes the number of tails, when a coin is tossed 6 times. Then,

X = difference between h and t = |h – t|

Now,             h : 0   1   2   3   4   5   6

Therefore,    t : 6   5   4   3   2    1   0

And, hence  X : 6   4   2   0   2   4   6

Therefore, the possible values of X are 6, 4, 2, 0.

53.  (i) The sample space of the random experiment ‘a coin is tossed twice’ is S = {HH, HT, TH, TT} ⇒ n(S) = 4

Let X denotes the random variable ‘number of heads’, then A can take the values 0, 1 or 2.

Probability distribution

(ii) Three coins tossed once = one coin tossed three times

∴ S = {(HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} ⇒ n(S) = 8

Let X denotes the random variable ‘number of heads’, then A can take the values 0, 1, 2 or 3.

Probability distribution

(iii) A coin is tossed four times = S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT} ⇒ n(S) = 16

Let X denotes the random variable ‘number of heads’, then A can take the values 0, 1, 2, 3 or 4.

Probability distribution

54.  S = {1, 2, 3, 4, 5, 6} = 6

(i) Let A be the set of favourable events. ⇒ n(A) = 1

(ii) Let A represents that 6 appears on one die ⇒ A = {6} ⇒ n(A) = 1

55.  n (S) = 30, A = {6 defective bulbs} ⇒n(A) = 6

n = 4 (4 bulbs are drawn with replacement), r = 0, 1, 2, 3, 4

56.  Let p represents the appearance of tail and represents the appearance of head.

Now q = 3p

57.  (i) Since, the sum of all the probabilities of a distribution is 1.

∴ P (X = 0) + P (X = 1) + ……. + P (X = 7) = 1

Since, k ≥ 0, therefore k = -1 is not possible.

(ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)

(iii) P (X > 6) = P (X = 7)

(iv) P (0 < X < 3)

58.  Probability distribution:

(a) P (X = 0) + P (X = 1) + P (X = 2) = 1

(b) P (X < 2) = P (X = 0) + P (X = 1)

59.  n(S) = 8

Let A denotes the appearance of head on a toss.

60.  Two dice thrown simultaneously is the same the die thrown 2 times.

Let S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6 x 6 = 36

Let A denotes the number 6 ⇒ A = {6} ⇒ n(A) = 1

61.  S = {(1, 2), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),

(1, 3), (2, 3), (3, 2), (4, 2), (5, 2), (6, 2),

(1, 4), (2, 4), (3, 4), (4, 3), (5, 3), (6, 3),

(1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (6, 4),

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}

n(S) = 30

Let X denotes the larger of the two numbers obtained.

62.  S = {(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),

(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4),

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}

n(S) = 36

Let A denotes the sum of the numbers = 2, B denotes the sum of the numbers = 3

C denotes the sum of the numbers = 4, D denotes the sum of the numbers = 5

E denotes the sum of the numbers = 6, F denotes the sum of the numbers = 7

G denotes the sum of the numbers = 8, H denotes the sum of the numbers = 9

I denotes the sum of the numbers = 10, J denotes the sum of the numbers = 11

K denotes the sum of the numbers = 12

63.  

64.  

65.  

66.  

67.  We know that the repeated throws of a die are Bernoulli’s triangles., Then

S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6

A = {1, 3, 5} ⇒ n(A) = 3

68.  n(S) = 36, A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} ⇒ n(A) = 6

69.  Let p = Probability of a success and q = Probability of a failure

70.  Let p = Probability of a success and q = Probability of a failure

S = 52 cards ⇒ n(S) = 52 and A = {13 spades} ⇒ n(A) = 13

71.  Let p = Probability of a success and q = Probability of a failure

p = 0.05 and q = 1 – 0.05 = 0.96

n = 5 and P (X = r) = C(n, r) prqn-r

(i) No bulb is fused, r = 0

(ii) Not more than one fused bulb

P (not more than one fused bulb) = P (X = 0) + P(X = 1)

(iii) P (more than one fused bulb out of 5) = 1 – [P (X = 0) + P (X = 1)]

= 1 – 1.2 (0.95)4        [BY (ii) part]

(iv) P (at least one fused bulb) = 1 – P(x = 0)

72.  S = {0, 1, 2 3, 4, 5, 6, 7, 8, 9} ⇒ n(S) = 10

Let A represents that the ball is marked with the digit 0.

Now P (one is marked with its digit 0) 

73.  S = {H, T} ⇒ n(S) = 2

Let A represents a head.

∴ A = {H} ⇒ n(A) = 1

74.  

xi = 0, 1, 2, 3, 4, 5, 6

P (X = r) = C (n, r)prqn-r

Since p = q

Therefore, P (X = r) = C(n, r) pn

Now, out of C (6, 0), C (6, 1), C (6, 2), C (6, 3), C (6, 4), C (6, 5), C (6, 6), here C (6, 3) is maximum.

Therefore, P (X = 3) is maximum, i.e., 

75.  

76.  

77.  

78.  S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6

Let A represents the favourable event i.e., 6

P (at most 2 success) = P (X = 0) + P (X = 1) + P (X = 2)

79.  

80.  

Therefore, option (C) is correct.

81.  

Therefore, option (A) is correct.

82.  A and B are two events such that P (A) ≠ 0

To find: P(B|A)

(i) A is a subset of B

83.  (i) Let B1 and G1 stand for male and female respectively.

Now the sample space is (S) = {B1B2, B1G2, B2G1, G1G2}

Let us consider the following events,

A = both are males

B = at least one is a male

∴ A = {B1B2} and B = {B1B2, B1G2, B2G1}

∴ Required probability 

(ii) Let A = both are females, A = {G1G2} and C = the older is a girl

∴ Required probability

84.  Men are represented by E1 and women are represented by E2.

A represents grey hair persons.

85.  

86.  S = {10 balls with mark X, 15 balls with mark Y} ⇒ n(S) = 25

Let a ball drawn with mark X be denoted by A.

A = {10 balls with mark X} ⇒ n(A) = 10

n = 6

(i) All will bear X mark, i.e., r = 6

(ii) Not more than 2 will bear mark Y.

∴ For Y mark, r = 0, 1, 2 and For X mark, r = 6, 5, 4

P (not more than 2 will bear mark Y) = P (X = 4) + P (X = 5) + P (X =6)

(iii) P (at least one ball will bear Y mark) = P (not more than 5 balls will bear mark X)

(iv) P (equal number of balls will bear mark X and Y) = P (3)

87.  p = Probability of knocking down a hurdle 

q = Probability of clearing a hurdle 

n = 10

P (He will knock down fewer than 2 hurdles) = P(0 ≤ x ≤ 2) = P (X = 0) + P (X = 1)

88.  S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6

A = {6} ⇒ n(A) = 1

Since the 3rd six is obtained in the 6th throw of die, there are two sixes i.e., two successes in the first 5 throws.

∴ Required probability = P (2 success in the first 5 throws) x P (success in the 6th throw)

89.  A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday then the year will have 53 Tuesdays.

Number of total days in a week = 7

Number of favourable days = 2

Therefore, P (the year will have 53 Tuesday) = 

90.  

91.  For one fair coin, probability of head  therefore, probability of tail 

Therefore, it is clear that n ≥ 4.

Since the minimum value of is 4, the man must loss the coin at least 4 times.

92.  When a die is thrown, probability of getting a six 

Therefore, 

(i) If he gets a six in first throw, then,

Probability of getting a six 

(ii) If he does not get a six, in first throw, but he gets a six in the second throw, then Its probability 

Probability that he does not get a six in first two throws and he gets a six in third throw 

Probability that he does not get a six in any of the three throws 

In first throw he gets a six, will receive Re.1

If he gets a six in second throw, he will receive Rs. (1 – 1) = 0

If he gets a six in third throw, he will receive Rs. (– 1 – 1 + 1) = Rs. – 1

= he will loss Rs. 1

If he does not get a six in all three throws, he will receive Rs. (–1 –1 –1) = Rs. –3

Expected value

93.  Let R represents the drawing of red ball and the four boxes are represented by A, B, C and D.

Since there are 4 bags.

94.  A patient has options to have the treatment of yoga and meditation and that of prescription of drugs.

Let these events be denoted by E1 and E2 i.e.,

E1 = Treatment of yoga and meditation

E2 = Treatment of prescription of certain drugs

Let A denotes that a person has heart attack, then P (A) = 40% = 0.40

Yoga and meditation reduces heart attack by 30.

⇒ Inspite of getting yoga and meditation treatment heart risk is 70% of 0.40

⇒ P(A|E1) = 0.40 x 0.70 = 0.28

Also, Drug prescription reduces the heart attack rick by 25%

Even after adopting the drug prescription hear rick is 75% of 0.40

95.  There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.

∴ Number of determinants that can be formed = 24 = 16

The value of determinants is positive in the following cases:

Therefore, the probability that the determinant is positive 

96.  Event A fails and B fails denoted by  respectively.

97.  Let E1 = Ball transferred from Bag I to Bag II is red

E2 = Ball transferred from Bag I to Bag Ii is black

A = Ball drawn from Bag II is red in colour

98.  A and B are two events such that P (A) ≠ 0 and P(B|A) = 1

Therefore, option (A) is correct.

99.  

Therefore, option (C) is correct.

100. P (A) + P (B) – P (A and B) = P (A)

Therefore, option (B) is correct.

MySchoolPage connects you with exceptional, certified maths tutors who help you stay focused, understand concepts better and score well in exams!

Have a Question?




Mathematics - Videos


Physics - Videos


Biology - Videos


Chemistry - Videos