CBSE Important Questions Class 11 Maths Chapter 14 – Probability

Class 11 Maths Chapter 14 Probability is part of the CBSE Class 11 Maths syllabus. It takes the ideas of probability further using the language of sets. Students learn about random experiments, the sample space, events as subsets of the sample space, and the different types of events such as complementary, mutually exclusive and exhaustive events. The chapter also gives the addition theorem of probability.

These CBSE important questions for Class 11 Maths Chapter 14 cover writing the sample space of an experiment, mutually exclusive and exhaustive events, the addition theorem for two events, probability with a deck of cards and with dice, and problems on at least one of two events happening. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define a random experiment, the sample space and an event. Write the sample space when two coins are tossed together.

Answer: A random experiment is an experiment whose outcome cannot be predicted with certainty, although all possible outcomes are known.
The sample space S is the set of all possible outcomes.
An event is any subset of the sample space.
For two coins tossed together:
S = {HH, HT, TH, TT}, so n(S) = 4.

Q2. A die is thrown. Describe the events (i) A: a number less than 4, (ii) B: a number greater than or equal to 4, (iii) A ∪ B and (iv) A ∩ B.

Answer: The sample space is S = {1, 2, 3, 4, 5, 6}.
(i) A = {1, 2, 3}
(ii) B = {4, 5, 6}
(iii) A ∪ B = {1, 2, 3, 4, 5, 6} = S, so A and B are exhaustive events.
(iv) A ∩ B = φ, so A and B are mutually exclusive events.
In fact B is the complement of A, written A′.

Q3. State the addition theorem of probability for two events. What does it become when the events are mutually exclusive?

Answer: For any two events A and B:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
This is the addition theorem.
If A and B are mutually exclusive, they cannot happen together, so A ∩ B = φ and P(A ∩ B) = 0.
In that case the theorem becomes
P(A ∪ B) = P(A) + P(B)

Q4. One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is (i) a king, (ii) a red card, (iii) a face card.

Answer: Total outcomes = 52
(i) King: There are 4 kings.
P = 4/52 = 1/13
(ii) Red card: There are 26 red cards (13 hearts and 13 diamonds).
P = 26/52 = 1/2
(iii) Face card: The face cards are the king, queen and jack of each of the 4 suits, so 12 in all.
P = 12/52 = 3/13

Q5. If P(A) = 0.42, P(B) = 0.48 and P(A ∩ B) = 0.16, find (i) P(not A), (ii) P(not B) and (iii) P(A or B).

Answer: (i) P(not A) = 1 − P(A) = 1 − 0.42 = 0.58
(ii) P(not B) = 1 − P(B) = 1 − 0.48 = 0.52
(iii) P(A or B) = P(A) + P(B) − P(A ∩ B)
= 0.42 + 0.48 − 0.16
= 0.74

Q6. Two dice are thrown together. Find the probability that the sum of the numbers is (i) 8 and (ii) at least 10.

Answer: Total outcomes = 6 × 6 = 36
(i) Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) — 5 outcomes.
P = 5/36
(ii) Sum at least 10 means the sum is 10, 11 or 12.
Sum 10: (4,6), (5,5), (6,4) — 3 outcomes
Sum 11: (5,6), (6,5) — 2 outcomes
Sum 12: (6,6) — 1 outcome
Total = 6 outcomes, so P = 6/36 = 1/6

Q7. Define mutually exclusive events and exhaustive events. Give one example of each.

Answer: Mutually exclusive events: Two events are mutually exclusive if they cannot occur at the same time, that is A ∩ B = φ.
Example: getting a head and getting a tail in a single toss of a coin.
Exhaustive events: A set of events is exhaustive if together they cover the whole sample space, that is their union is S.
Example: when a die is thrown, the events "an even number" and "an odd number" together cover all six outcomes.

Q8. A letter is chosen at random from the word ASSASSINATION. Find the probability that the letter is (i) a vowel and (ii) a consonant.

Answer: The word ASSASSINATION has 13 letters.
Vowels: A, A, I, A, I, O — that is 6 vowels.
Consonants: 13 − 6 = 7
(i) P(vowel) = 6/13
(ii) P(consonant) = 7/13
(Note that the two probabilities add up to 1, as they must.)

Q9. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both. If one student is selected at random, find the probability that the student has opted for NCC or NSS.

Answer: Let A be the event of opting for NCC and B for NSS.
P(A) = 30/60 = 1/2
P(B) = 32/60 = 8/15
P(A ∩ B) = 24/60 = 2/5
Using the addition theorem:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 1/2 + 8/15 − 2/5
Taking the LCM 30: = (15 + 16 − 12)/30
= 19/30
So the probability is 19/30.

Q10. Three coins are tossed once. Find the probability of getting (i) exactly two heads, (ii) at least two heads and (iii) no head.

Answer: Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, so n(S) = 8.
(i) Exactly two heads: HHT, HTH, THH — 3 outcomes.
P = 3/8
(ii) At least two heads means two or three heads: HHT, HTH, THH, HHH — 4 outcomes.
P = 4/8 = 1/2
(iii) No head means all tails: TTT — 1 outcome.
P = 1/8

Q11. If the probability that a student passes in Mathematics is 2/3 and in English is 4/9, and the probability of passing in both is 1/4, find the probability of passing in at least one subject.

Answer: Let M be passing in Mathematics and E in English.
P(M) = 2/3, P(E) = 4/9 and P(M ∩ E) = 1/4
"At least one subject" means M ∪ E:
P(M ∪ E) = P(M) + P(E) − P(M ∩ E)
= 2/3 + 4/9 − 1/4
Taking the LCM 36: = (24 + 16 − 9)/36
= 31/36
So the probability is 31/36.

Q12. A bag contains 9 discs, of which 4 are red, 3 are blue and 2 are yellow. One disc is drawn at random. Find the probability that it is (i) not blue and (ii) either red or blue.

Answer: Total discs = 9
(i) Not blue: P(blue) = 3/9 = 1/3
P(not blue) = 1 − 1/3 = 2/3
(ii) Either red or blue: These are mutually exclusive, so we simply add.
P(red or blue) = 4/9 + 3/9
= 7/9
So the probability is 7/9.

Related Study Material on Extramarks

Q.1 Which of the following are mathematically acceptable statements?
(i) All irrational numbers are real numbers.
(ii) How wonderful
(iii) In 2004 the president of USA was a woman.
(iv) What are you doing?

Marks:4

Ans

(i) All irrational numbers are real numbers is a statement because it is true.
(ii) How wonderful! It is not a statement because it is an exclamation.
(iii) In 2004, the president of USA was a woman is a statement because it is false.
(iv) What are you doing? It is not a statement because it is a question.

Q.2 Verify by the method of contradiction:p: ?3 is irrational.

Marks:4

Ans

In this method, we assume that the given statement is false, i.e., we assume that 3 is a rational number. This means that there exists positive integers a and b such that 3 = a/b, where a and b have no common factors.
Squaring the equation, we get
3 = a2/b23b2 = a2 3 divides a.
Therefore, there exists an integer c such that a = 3c.
Then, a2 = 9c2 and 3 b2 = a2
Hence, 3b2 = 9c2 ⇒ b2 = 3c2 3 divides b. But, we have already shown that 3 divides a. It means 3 is a common factor of a and b. This shows that our assumption is wrong. Thus, the statement 3 is irrational is true.

Q.3 Write the contrapositive of the following statement:

(i) If a number is divisible by 4 then it is divisible by 2.
(ii) If you are born in Nepal then you are a citizen of Nepal.
(iii) If a number n is odd then n2 is odd.
(iv) Something is hot implies that it has a high temperature.

Marks:4

Ans

(i) If a number is not divisible by 2 then it is not divisible by 4.
(ii) If you are not a citizen of Nepal then you were not born in Nepal.
(iii) If n2 is not even, then n is not even.
(iv) If something has not high temperature, then it is not hot.

Q.4 Find the component statements of the following compound statements:

(i) The roof is white and the wall is red.
(ii) 0 is a positive number or a negative number.

Marks:2

Ans

(i) The component statements are:
p: The roof is white.
q: The wall is red.

(ii) The component statements are:
p: 0 is a positive number.
q: 0 is a negative number.

Q.5 Show by the method of contradiction that the following statement is true:p: If x is a real number such that x5 + 4x = 0,  then x is 0.

Marks:4

Ans

Let q and r be the statements given by
q : x is a real number such that x5 + 4x = 0
r: x is 0
Then, p: if q, then r.
If possible, let p be not true.
Then p is true
(q r) is true.
q and r is true.
x is a real number such that x5 + 4x = 0 and x  0
x = 0 and x  0
This is a contradiction.
Hence, p is true.

Please register to view this section