Probability Class 11 Mathematics Important Questions

Please refer to Probability Class 11 Mathematics Important Questions with solutions provided below. These questions and answers have been provided for Class 11 Mathematics based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 11 Mathematics for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 11.

Class 11 Mathematics Important Questions Probability

Assertion & Reasoning Based MCQs :

(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement..

Question. Consider the experiment of rolling a die. Then, sample space is S = { 1, 2, 3, 4, 5, 6}
Assertion : The event E : “the number appears on the die is a multiple of 7”, is an impossible event.
Reason : The event F : “the number turns up is odd or even”, is a sure event.

Question. Assertion : If sample space of an experiment is S = {1, 2, 3, 4, 5, 6} and the events A and B are defined as
A : “a number less than or equal to 3 appears”
B : “a number greater than or equal to 3 appears”, then A and B are exhaustive events.
Reason : Events are exhaustive if atleast one of them necessarily occur whenever the experiment is performed.

Question. Assertion : The probability of drawing either an ace or a king from a pack of cards in a single draw is 2/13.
Reason : For two events A and B which are not mutually exclusive, P(A ∪ B) = P(A) + P(B) – P(A ∩ B).

Question. Assertion : A coin is tossed. If it shows head, we draw a ball from a bag consisting of 3 brown and 4 red balls; if it shows tail we throw a die, then the sample space of this experiment is S = {HB1, HB2, HB3, HR1, HR2, HR3, HR4, T1, T2, T3, T4, T5, T6}
Reason : Consider the experiment in which a coin is tossed repeatedly until a head comes up, then the sample space is S = {H, TH, TTH, TTTH,……..}

Question. Assertion : A coin is tossed and then a die is rolled only in case a head is shown on the coin.
The sample space for the experiment is S = {H1, H2, H3, H4, H5, H6, T}
Reason : 2 boys and 2 girls are in room X, and 1 boy and 3 girls are in room Y. Then, the sample space for the experiment in which a room is selected and then a person, is S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4, YG5} where Bi, denote the boys and Gj, denote the girls.

Very Short Answer Type Questions :

Question. Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.
Answer: The sample space S for selecting three bulbs at random from a lot is given by
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN }
where D indicates a defective bulb and N a non-defective bulb.

Question. If A and B are mutually exclusive events of a random experiment and P(A∪B) = 0.75, P(A) = 0.6, find P(B).
Answer: Given, P(A ∪ B) = 0.75 and P(A) = 0.6
Now, P(A) = 1 − P(A) ⇒ 0.6 = 1 – P(A)
⇒ P(A) = 0.4
Also, P(A ∪ B) = P(A) + P(B) ⇒ 0.75 = 0.4 + P(B)
(Q A and B are mutually exclusive events)
⇒ P(B) = 0.75 – 0.4
∴ P(B) = 0.35

Question. Given two mutually exclusive events A and B such that P(A) = 1/2 and P(B) = 1/3 , find P (A ∪ B)′.
Answer: Given, P(A) = 1/2 , P(B) = 1/3
∴ P(A ∪ B) = P(A) + P (B) = 1/2 + 1/3 = 3+2/6 = 5/6
P(A ∪ B)′ = 1 – P(A ∪ B) = 1 -5/6 = 1/6

Question. The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.
Answer: The sample space S for the given experiment is
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}.

Question. An experiment consists of tossing a coin once and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.
Answer: The sample space S for given experiment is :
S = {HH, HT, T1, T2, T3, T4, T5, T6}

Question. If P (E′ ∩ F′) = 0.87, then find P(E ∪ F).
Answer: Given, P(E ′ ∩ F ′ ) = 0.87
⇒ P(E ′ ∩ F ′ ) = 1 – P(E ∪ F) ⇒ 0.87 = 1 – P(E ∪ F)
⇒ P(E ∪ F) = 1 – 0.87 = 0.13

Question. If E and F are events such that P(E) = 1/4 .
P(F) = 1/2 and P(E and F) = 1/8 , find P(E or F).
Answer: Given, P(E)= 1/4 , P(F) = 1/2 , P(E ∩ F) 1/8
Now, P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
⇒ P(E ∪ F) = 1/4 + 1/2 – 1/8 = 2+4-1/8
∴ P(E ∪ F) = 5/8

Question. If P(A) = 3/5 and P(B) = 1/5 , find P(A or B), if A and B are mutually exclusive events.
Answer: Given, P(A) = 3/5 , P(B) = 1/5
Now, P(A ∪ B) = P(A) + P(B) = 3/5 + 1/5
[Q A and B are mutually exclusive events]
∴ P(A or B) = P(A ∪ B) = 4/5

Question. A coin is tossed once. If it shows head, we draw a ball from a bag consisting of 3 blue and 4 white balls. If it shows tail, we throw a die twice. Describe sample space.
Answer: When a coin is tossed and it shows head, then
sample space, S1 = {HB1, HB2, HB3, HW1, HW2, HW3, HW4}
When the coin shows tail (T) then a die is thrown twice.
S2 = {T(1, 1), T(1, 2), ….., T(1, 6), T(2, 1), T(2, 2),…., T(2,
6), T(3, 1), ….., T(3, 6), ….., T(6, 1), ….., T(6, 6)}
Therefore, S = {HB1, HB2, HB3, HW1, HW2, HW3, HW4,
T(1,1), T(1,2), …..,T(1,6), T(2,1), ….., T(2,6), T(3,1), ….., T(3,6),
T(4,1), ….., T(4,6), T(5,1), ….., T(5,6), T(6,1), ….., T(6,6)}

Question. The probability that at least one of the events E1 and E2 occurs is 0.6. If the probability of the simultaneous occurrence of E1 and E2 is 0.2, find P(E1) + P(E2).
Answer: Given, P(E1 ∪ E₂) = 0.6 and P(E₁ ∩ E₂) = 0.2
∴ P(E1 ∪ E2) = P(E₁) + P(E₂) –P(E1 ∩ E₂)
⇒ P(E₁) + P(E₂) = P(E₁ ∪ E₂) + P(E₁ ∩ E₂)
= (0.6 + 0.2) = 0.8
⇒ P(E₁) + P(E₂) = 0.8
⇒ {1 – P(E–1)} + {1 – P(E–2)} = 0.8
⇒ P(E–1) + P(E–2) = (2 – 0.8) = 1.2
Hence, P(E–1) + P(E–2) = 1.2

Short Answer Type Questions :

Question. Two dice are thrown simultaneously. Let E1 denote getting a doublet, E2 denote getting sum of the numbers appearing on the dice to be at least 10.
(i) Find P(E1 or E₂).
(ii) Are E1 and E2 mutually exclusive?
Answer: Total number of possible outcomes = 36
E₁ = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
E₂ = {(6, 4), (4, 6), (5, 5), (6, 5), (5, 6), (6, 6)}
∴ P(E1) = 6/36 , P(E2 ) = 6/36
and P(E₁ ∩ E₂ ) = 2/36 .
(i) P(E₁ or E₂) = P(E₁ ∪ E₂)
= P(E₁) + P(E₂) – P(E1 ∩ E₂)
= 6/36 + 6/36 – 2/36 = 6+6-2/36 = 5/18
(ii) Q E₁ ∩ E₂ = {(5, 5), (6, 6)} ≠ φ
∴ E₁ and E₂ are not mutually exclusive.

Question. If a card is drawn from a deck of 52 cards, then find the probability of getting a king or a heart or a red card.
Answer: Let S be the sample space \ n(S) = 52
Let A, B, C be the events of getting a king, a heart and a red card respectively.
∴ P(A) = 4/52 = 1/13 , P(B) = 13/52 = 1/4 , P(C) = 26/52 = 1/2
P(A∩B) = 1/52 , P(A∩C) = 2/52 = 1/26 , 
P(B∩C) = 13/52 = 1/4 and P(A∩B∩C) = 1/52
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B)
– P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= 1/13 + 1/4 + 1/2 – 1/52 – 1/4 – 1/26 + 1/52 = 7/13
∴ Probability of getting a king or a heart or a red card = 7/13 .

Question. A die is rolled. Let ‘E ‘ be the event “die shows prime number” and ‘F ‘ be the event “die shows even number”. Are E and F mutually exclusive?
Answer: E = {2, 3, 5} and F = {2, 4, 6}
∴ E ∩ F = {2} ≠ φ
∴ E and F are not mutually exclusive.

Question. Probability that a truck stopped at a roadblock will have faulty brakes or badly worn tires are 0.23 and 0.24 respectively. Also, the probability is 0.38 that a truck stopped at the roadblock will have faulty brakes or badly working tires. What is the probability that a truck stopped at this roadblock will have faulty brakes as well as badly worn tires?
Answer: Let B be the event that a truck stopped at the roadblock will have faulty brakes and T be the event that it will have badly worn tires.
Given, P(B) = 0.23, P(T) = 0.24 and P(B ∪ T ) = 0.38.
We have to find P(B ∩ T).
Now, P(B ∪ T ) = P (B) + P(T) – P(B ∩ T)
[By addition theorem]
⇒ P(B ∩ T ) = P (B) + P(T) – P(B ∪ T)
= 0.23 + 0.24 – 0.38 = 0.09

Question. For a post, three persons A, B and C appear in the interview. The probability of A being selected is twice that of B and the probability of B being selected is thrice that of C. What are the individual probabilities of A, B, C being selected?
Answer: Let A₁, A₂ and A₃ be three events as defined below:
A1 = Person A is selected, A2 = Person B is selected, A₃
= Persons C is selected.
We have, P(A1) = 2P (A2) and P(A2) = 3P (A3)
⇒ P(A1) = 6P (A3) and P(A2) = 3 P(A3).
Since A1, A2, A3 are mutually exclusive and exhaustive
events.
∴ A1 ∪ A2 ∪ A3 = S
⇒ P(A1) + P(A2) + P(A3) = 1
⇒ 6P(A3) + 3 P(A3) + P(A3) = 1
⇒ 10P(A3) = 1
⇒ P(A3) = 1/10 
∴ P(A1) = 6/10 and P(A2) = 3/10

Question. In an essay competition, the odds in favour of competitors P, Q, R, S are 1 : 2, 1 : 3,1 : 4 and 1 : 5 respectively. Find the probability that one of them wins the competition.
Answer: Let A, B, C, D be the events that the competitors P, Q, R and S respectively win the competition. Then, P(A) = 1/3 , ,P(B) = 1/4 , P(C) = 1/5 and P(D) = 1/6
Since only one competitor can win the competition.
Therefore, A, B ,C, D are mutually exclusive events.
∴ Required probability = P(A ∪ B ∪ C ∪ D)
= P(A) + P(B) + P(C) + P(D) [By addition theorem]
= 1/3 + 1/4 + 1/5 + 1/6 = 19/20

Question. A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?
11. The sample space S for the experiment is given by
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41,
H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}.

Question. If P(A) = 0.59, P(B) = 0.30 and P(A ∩ B) = 0.21, then find P(A′ ∩ B′).
Answer: P(A′ ∩ B′) = P(A ∪ B)′ = 1 – P(A ∪ B)
= 1 – P(A) – P(B) + P(A ∩ B) = 1– 0.59 – 0.30 + 0.21= 0.32

Question. A coin whose faces are marked by 3, 4 is tossed 5 times. Find the probability of getting a total of 24.
Answer: When we toss a coin five times and denote the sum by S, then 15 ≤ S ≤ 20 which means we never get the sum 24
∴ Required probability is 0.

Question. From a group of 2 boys and 3 girls, two children are selected at random. Consider the following events:
(i) A : Event that both the selected children are girls
(ii) B : Event that the selected group consists of one boy and one girl
(iii) C : Event that at least one boy is selected Which pairs of events are mutually exclusive?
Answer: Let us name the boys as B1 and B2, and the girls as G₁, G₂ and G₃.
Then, S = {B1B2, B1G1, B₁G₂, B₁G₃, B₂G1, B₂G₂, B₂G₃, G₁G₂, G₁G₃, G₂G₃}.
We have, (i) A = {G₁G₂, G₁G₃, G₂G₃}
(ii) B = {B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃}
(iii) C = {B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃, B₁B₂}
Clearly, A ∩ B = φ and A ∩ C = φ.
Hence, (A, B) and (A, C) are mutually exclusive events.

Question. In a lot of 12 Microwave ovens, there are 3 defective units. A person has ordered 4 of these units and since each is identically packed, the selection will be random. What is the probability that (i) all 4 units are good (ii) exactly 3 units are good (iii) at least 2 units are good?
Answer: Total number of possible outcomes = ¹²C₄
Number of defective units = 3
Number of units that are good = 9
(i) Number of ways of selecting all 4 good units = 9C4
So, required probability = ⁹C₄/¹²C₄ = 126/495 = 14/55 .
(ii) Number of ways of selecting exactly 3 good units = ⁹C₃ x ³C₁
∴ Required probability = 9C3 x 3C1/12C4 = 84×3/495 = 28/55
(iii) Number of favourable events in which atleast 2 units are good = ⁹C₂ × ³C₂ + ⁹C₃ × ³C₁ + ⁹C₄
Required probability = ⁹C₂ . ³C₂ + ⁹C₃ . ³C₁ + ⁹C₄/¹²C₄
= 108 + 252 + 126 /495 = 486/495 = 54/55

Question. A coin is tossed three times, consider the following events :
A : ‘No head appears’
B : ‘Exactly one head appears’
C : ‘At least two heads appear’
Do they form a set of mutually exclusive and exhaustive events?
Answer: If S be the sample space of tossing a coin three times
then S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }
Now, A (Event when no head appears) = {TTT }
B (Event when exactly one head appears)
= {HTT, THT, TTH }
C (Event when atleast two heads appear)
= {HHT, HHH, HTH, THH }
Clearly, A ∩ B ∩ C = φ
So, A, B and C are mutually exclusive events
Clearly, A ∪ B ∪ C = S
So, A, B and C are exhaustive events.

Question. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that (i) A is a simple event
(ii) B and C are compound events
(iii) A and B are mutually exclusive events.
Answer: We have, A = {(1, 1)}, B = {(1, 2), (2, 1)}
and C = {(1, 3), (3, 1), (2, 2)}
(i) Since A consists of a single sample point, it is a simple event.
(ii) Since both B and C contain more than one sample point, hence each one of them is a compound event.
(iii) Since A ∩ B = φ,
∴ A and B are mutually exclusive events.

Question. What is the probability that in a group of (i) 2 people, both will have the same birthday?
(ii) 3 people, at least two will have the same birthday?
(Assuming that there are 365 days in a year.)
Answer: (i) First person may have any one of the 365 days of the year as a birthday.
Similarly, second person may have any one of 365 days of the year as a birthday.
So, the total number of ways in which two persons may
have their birthdays = 365 × 365 = (365)2
The number of ways in which two persons have the same birthday = 365.
Hence, required probability = 365/365² = 1/365
(ii) Let A be the event “At least two people have the same birthday”. Then, A – = No two or more people have the same birthday = All the three persons have distinct birthdays.
∴ P(A) = 365 x 364 x 363 /365³ = 364 x 363/365²
Hence, required probability = 1 – P(A ^–) 
= 1 − 364 x 363/365²

Question. If E and F are two events such that P(E) = 1/4 , P(F) = 1/2 and P(E and F)= 1/8 , find (i) P(E but not F) (ii) P(F but not E)
Answer: Given, P(E) = 1/4, P(F) = 1/2and P(E and F) = 1/8 ⇒ P(E∩ F) = 1/8
(i) P(E but not F ) = P(E ∩ F–) = P(E ) –P(E ∩ F)
= 1/4 – 1/8 = 2-1/8 = 1/8 .
(ii) P(F but not E ) = P(F ∩ E–) = P(F ) –P(F ∩ E)
= 1/2 – 1/8 = 4-1/8 = 3/8 .

Question. (i) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 4?
(ii) What is the probability that the sum of the numbers on the two faces is divisible by 3 or 4?
Answer: Let S be the sample space associated with the
experiment of throwing a pair of dice.
Then, n(S) = 36.
∴ Total number of possible outcomes = 36
Consider the following events.
A : The sum of the numbers on two faces is divisible by 3
B : The sum of the numbers on two faces is divisible by 4.
Then, A = {(1, 2), (2, 1), (1, 5), (5, 1), (3, 3), (2, 4), (4, 2),
(3, 6), (6, 3), (4, 5), (5, 4), (6, 6)}
B = {(2, 2), (1, 3), (3, 1), (2, 6), (6, 2), (4, 4), (3, 5), (5, 3), (6, 6)} and A ∩ B ={(6, 6)}
∴ P(A) = 12/36 = 1/3 , P(B) = 9/36 = 1/4 and P(A∩B) = 1/36 
(i) Required probability = P(¯A∩¯B) = P(¯A∩¯B)
= 1 – P (A ∪ B) = 1 – {P(A) + P(B) – P(A ∩ B)}
= 1-{1/3 + 1/4 – 1/36} = 4/9
(ii) Required probability = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B) = 1/3 + 1/4 – 1/36 = 5/9 .

Question. A five digit number is formed by the digits 1, 2, 3, 4, 5 without repetition. Find the probability that the number is divisible by 4.
Answer: Total number of five digit numbers formed by the digits 1, 2, 3, 4, 5 is 5!
∴ Total number of possible outcomes = 5! = 120.
We know that a number is divisible by 4 if the number formed by last two digits is divisible by 4. Therefore last two digits can be 12, 24, 32, 52
that is, last two digits can be filled in 4 ways. But corresponding to each of these ways there are 3! = 6 ways of filling the remaining three places.
Therefore the total number of five digit numbers formed by the digits 1, 2, 3, 4, 5 and divisible by 4 = 4 × 6 = 24
∴ Favourable number of elementary events = 24
So, required probability = 24/120 = 1/5

Question. The probability of two events A and B are 0.25 and 0.50 respectively. The probability of their simultaneous occurrence is 0.14. Find the probability that neither A nor B occurs.
Answer: We have, P(A) = 0.25, P(B) = 0.50 and
P (A ∩ B) = 0.14
∴ Required probability = P(A B) = P(A B)
= 1 – P (A ∪ B) = 1 – [P(A)+ P(B) – P(A ∩ B)]
= 1 – (0.25 + 0.50 – 0.14) = 0.39

Question. P and Q are two candidates seeking admission in I.I.T. The probability that P is selected is 0.5 and the probability that both P and Q are selected is atmost 0.3. Prove that the probability of Q being selected is atmost 0.8.
Answer: Let A1 and A2 be two events defined as:
A1 = P is selected, A2 = Q is selected.
We have, P(A1) = 0.5 and P(A1 ∩ A2) ≤ 0.3
Now, P(A1 ∪ A2) ≤ 1
⇒ P(A1) + P(A2) – P(A1 ∩ A2) ≤ 1
⇒ 0.5 + P(A2) – P(A1 ∩ A2) ≤ 1
⇒ P(A2) ≤ 0.5 + P(A1 ∩ A2)
⇒ P(A2) ≤ 0.5 + 0.3 ⇒ P(A2) ≤ 0.8

Question. Four boys and two girls sit in a row at random. Find the probability that two girls do not sit together.
Answer: Consider 2 girls as a single unit. Then possible
arrangements such that 2 girls sit together = 5!
= 120
The number of required ways in which 2 girls are never
sit together = 6! – (5! × 2)
= 720 – 240 = 480
∴ Required probability = 480/6! = 480/720 = 2/3

Question. Consider the event A = {2, 4, 6, 8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly, the sample space is S = {1, 2, 3, ……., 10}. Then, which of the following is/are true?
(i) P(A′) = 1 – P(A) (ii) P(A’) = 3/5
(iii) P(A) + P(A′) = 1
Answer: If all the outcomes 1, 2, ……., 10 are considered to be equally likely, then the probability of each outcome is 1/10 .
Now, P(A) = P(2) + P(4) + P(6) + P(8)
= 1/10 + 1/10 + 1/10 + 1/10 = 4/10 = 2/5
Also event (not A) = A′ = {1, 3, 5, 7, 9, 10}
∴ P(A′) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10)
= 6/10 = 3/5
Also, we know that A′ and A are mutually exclusive and exhaustive events i.e.,
A ∩ A′ = f and A ∪ A′ = S or P(A ∪ A′) = P(S)
⇒ P(A) + P(A′) = 1
⇒ P(A′) = P(not A) = 1 – P(A)

Question. A drawer has 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or a bolt?
Answer: Let A be the event that the item chosen is rusted and B be the event that the item chosen is a bolt.
Total number of possible outcomes = 200
As, half of nuts and bolts are rusted
∴ Number of rusted items = 25 + 75 = 100
Hence, P(A) = 100/200 , P(B) = 50/200
and P(A ∩ B) = 25/200
∴ Required probability = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= 100/200 + 50/200 – 25/200 = 125/200 = 5/8 .

Question. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination? Explain briefly the importance of examination.
Answer: Let E represents the event that student passes the English examination and H represents the event that student passes the Hindi examination.
Given, P(E ∩ H) = 0.5 , P(E– ∩ H – ) = 0.1, P(E) = 0.75
Now, P(E– ∩ H – ) = 1 – P (E ∪ H )
⇒ 0.1 = 1 – [P(E) + P(H) – P(E ∩ H)]
⇒ 0.1 = 1 – [0.75 + P(H) – 0.5]
⇒ P(H) = 1 – 0.75 + 0.5 – 0.1 = 1.5 – 0.85 = 0.65
Therefore, the probability of passing the Hindi examination is 0.65.

Question. What is the probability that all L’s come together in the word PARALLEL?
Answer: Total number of possible outcomes = 8!/3!2! = 3360
Total number of favourable outcomes (3L’s as one letter) =6!/2! = 360
Hence, required probability = 360/3360 = 3/28

Question. If E and F are events such that P(E) = 1/4 , P(F) = 1/2 and P(E and F) = 1/8 , (i) P(E or F) (ii) P(not E and not F).
Answer: Given, P(E) = 1/4 , P(F) = 1/2 and P(E ∩ F) = 1/8
(i) P(E or F) = P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
= 1/4 + 1/2 – 1/8 = 2+4-1/8 = 5/8
(ii) P(not E and not F) = P(E′ ∩ F′) = P(E ∪ F)′
= 1 – P(E ∪ F) = 1 −5/8 = 3/8

Long Answer Type Questions :

Question. The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will not get an electric contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that he will get both?
Answer: Let A be the event that a contractor will get a plumbing contract and B be the event that a contractor will get an electric contract.
P(A) = 2/3 , ,P(B) = 5/9 , P(A∪B) = 4/5
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 4/5 = 2/3 + [1-P(B)] – P(A ∩ B)
⇒ 4/5 = 2/3 + 1 – 5/9 – (A ∩ B)
⇒ P(A∩B) =2/3 + 1-5/9 – 4/5
= 30 + 45 – 25 – 36/45 = 14/15

Question. In a single throw of two dice, determine the probability of obtaining a total of 7 or 9.
Answer: Total number of events = 36
Let A be the event of getting sum of 7
= {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Let B be the event of getting sum of 9
= {(3, 6), (4, 5), (5, 4), (6, 3)}
P(A) = 6/36 = 1/6 , P(B) = 4/36 = 1/9 , P(A∩B) = φ 
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 1/6 + 1/9 = 3+2/18 = 5/18

Question. 20 cards are numbered from 1 to 20. One card is then drawn at random. What is the probability that the number on the card drawn is
(i) a prime number?
(ii) an odd number?
(iii) a multiple of 5?
(iv) not divisible by 3?
Answer: Clearly, the sample space is given by
S = {1, 2, 3, 4, 5, …., 19, 20} ∴ n(S) = 20.
(i) Let E1 denotes the event of getting a prime number.
Then, E1 = {2, 3, 5, 7, 11, 13, 17, 19}
∴ n(E1) = 8.

Question. An experiment involves rolling a pair of dice and recording the numbers that come up.
Describe the following events :
A : the sum is greater than 8
B : 2 occurs on either die
C : the sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?
Answer: When two dice are thrown, there are
6 × 6 = 36 possible outcomes.
∴ S = {(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), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3),
(6, 4), (6, 5), (6, 6)}
∴ A = The sum is greater than 8
= {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3),
(6, 4), (6, 5), (6, 6)}
B = 2 occurs on either die
= {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 2), (4, 2), (5, 2), (6, 2)}
C = The sum is at least 7 and a multiple of 3
= {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
Also, A ∩ B = f, B ∩ C = f
and A ∩ C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
(i) Since A ∩ B = f. So A and B are mutually exclusive.
(ii) Since B ∩ C = f. So B and C are mutually exclusive.
(iii) Since A ∩ C ≠ f. So A and C are not mutually exclusive.

Question. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(a) the student opted for NCC or NSS.
(b) the student has opted neither NCC nor NSS.
(c) the student has opted NSS but not NCC.
Answer: Let A represents the event that student opted for
NCC and B represents the event that student opted for NSS.
Given, n(A) = 30, n(B) = 32, n(A ∩ B) = 24
(a) P(Student opted for NCC or NSS) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 30/60 + 32/60 – 24/60 = 30+32-24/60 = 38/60 = 19/30
(b) Probability for students that opted neither NCC nor
NSS = P(A′ ∩ B′) = P(A ∪ B)′ = 1 – P(A ∪ B)
= 1 – 19/30 = 11/30
Probability that students opted for NSS but not NCC
= P(A ∩ B) = P(B) − P(A∩B)
= 32/60 – 24/60 = 8/60 = 2/15

Case Based Questions :

Case I : Read the following passage and answer the questions from 36 to 40.
In Richa’s home, She, Suhani and Reetika who are best friends are playing with colourful discs which are similar in shape and size. Richa has 4 red discs, Suhani has 3 blue discs and Reetika has 2 yellow discs. They put all the 9 discs in a bag.
Suhani ask Richi’s mother to drawn any discs from the bag randomly.

Question. Probability that the drawn disc is not blue, is
(a) 1/3
(b) 2/9
(c) 2/3
(d) 5/6

Question. The probability that the drawn disc is either red or blue, is
(a) 7/9
(b) 1/3
(c) 3/22
(d) 7/13

Question. Find the probability that the drawn disc to be red.
(a) 1/3
(b) 4/9
(c) 5/9
(d) 7/13

Question. Find the probability that the drawn disc is yellow.
(a) 2/9
(b) 4/9
(c) 3/9
(d) 2/13

Question. Which disc has more probability to be drawn?
(a) yellow
(b) orange
(c) blue
(d) red

Case II : Read the following passage and answer the questions from 41 to 45.
Nishant has an electricity shop. From an agent he got three bulbs which are manufactured by a new company. Nishant wants to know if any of three bulb is defective and classified as Good ‘non-defective’ and bad ‘defective’.

Question. The probability that there are atleast two defective bulb is
(a) 9/32
(b) 1/2
(c) 5/9
(d) 7/13

Question. The probability that all the bulbs are good is
(a) 1/8
(b) 5/8
(c) 2/8
(d) 3/8

Question. What is the sample space of three bulb to be good or bad?
(a) {BBB, BBG, BGB, GBB, GBG, BGG, GGG}
(b) {BBB, BGG, GBG, GGB, GGG}
(c) {BBB, BBG, BGG, GGG}
(d) {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}

Question. Find the probability that there are no defective bulb?
(a) 1/8
(b) 2/8
(c) 3/8
(d) 5/8

Question. What is the probability that there is exactly one bad bulb?
(a) 7/8
(b) 5/8
(c) 3/8
(d) 6/8

Case III : Read the following passage and answer the questions from 46 to 50.
A management committee of a residential colony decided to award two members which will be selected from two men and two women for honesty, helping others and for supervising the workers to keep the colony neat and clean.

Question. The probability that the committee will select 1 man is
(a) 9/10
(b) 5/6
(c) 2/3
(d) 1/3

Question. The probability that the committee will select at most 1 man is
(a) 1/6
(b) 1/3
(c) 2/3
(d) 5/6

Question. The probability that the committee will select exactly 2 men is
(a) 1/6
(b) 5/6
(c) 3/8
(d) 7/9

Question. The total number of ways in which any of two members will be selected is
(a) 4
(b) 5
(c) 6
(d) 7

Question. The probability that the committee will select no man is
(a) 2/6
(b) 1/6
(c) 4/6
(d) 5/6