Probability — Class 11 MCQs with Answers

Probability — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Probability of head on a fair coin:

   A1/4B1C1/2D0
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   Answer: 1/2

   Explanation: A fair coin has 2 equally likely outcomes, so P(head) = 1/2. Picking 1/4 comes from wrongly treating this as a two-coin toss with 4 outcomes instead of one coin with 2.

2. Q2 of 6Easy

   Probability P(A) lies in:

   ARB[0,1]C[−1,1]D(0,∞)
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   Answer: [0,1]

   Explanation: Probability is always a value between 0 and 1 inclusive, since it's a fraction of favourable to total outcomes — never negative or above 1. [−1,1] confuses it with a correlation-type range.

3. Q3 of 6Medium

   When two dice are thrown together, the number of elements in the sample space is:

   A24B12C6D36
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   Answer: 36

   Explanation: Each die has six outcomes, giving 6 × 6 = 36 ordered pairs.

4. Q4 of 6Medium

   Assertion (A): When a die is rolled, the events “an odd number appears” and “a number less than 4 appears” are not mutually exclusive. Reason (R): A sample space containing n distinct elements has exactly n simple events.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true but R is not the correct explanation of A

   Explanation: A is true: {1, 3, 5} ∩ {1, 2, 3} = {1, 3} ≠ φ. R is also true, but it is about counting simple events and does not explain why these two events overlap.

5. Q5 of 6Hard

   P(drawing 3 cards: all kings) from a deck:

   AC(4,3)/C(52,3) = 4/22100B1/52³C4/52³D12/52
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   Answer: C(4,3)/C(52,3) = 4/22100

   Explanation: Choosing 3 kings out of the 4 available, over all C(52,3) possible 3-card hands, gives P = C(4,3)/C(52,3) = 4/22100. Using 52³ in the denominator instead wrongly assumes order and repetition matter.

6. Q6 of 6Hard

   Assertion (A): If P(E) = 1/4, P(F) = 1/2 and P(E and F) = 1/8, then P(not E and not F) = 3/8. Reason (R): If two events are mutually exclusive, the probability that both occur is 0.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true but R is not the correct explanation of A

   Explanation: P(E ∪ F) = 1/4 + 1/2 – 1/8 = 5/8, and “not E and not F” = (E ∪ F)′, so its probability is 1 – 5/8 = 3/8; A is true. R is true, but E and F are not mutually exclusive here (P(E and F) = 1/8), so R does not explain A.

Probability notes: key concepts (Class 11 Mathematics)

This chapter puts chance on a firm axiomatic footing. Starting from random experiments, sample spaces and events as subsets of outcomes, it lays down the axioms of probability, recovers the classical favourable-over-total rule for equally likely outcomes, and proves the addition theorem for the union of two events.

Random experiment

An experiment whose set of possible outcomes is known but whose actual outcome cannot be predicted in advance, such as tossing a coin.

Sample space

The set S of every possible outcome of a random experiment; each individual outcome in S is called a sample point.

Event

Any subset of the sample space; an event occurs when the actual outcome of the experiment belongs to that subset.

Simple and compound events

A simple (elementary) event has a single sample point; a compound event contains two or more sample points of S.

Mutually exclusive events

Events that cannot occur together, so their intersection is empty (A∩B = ∅); the occurrence of one rules out the other.

Exhaustive events

A set of events whose union is the whole sample space, so at least one of them must occur on every trial of the experiment.

Complementary event

The complement A′ ('not A') is every outcome of S not in A; A and A′ are mutually exclusive and together exhaustive.

Axioms of probability

P(E) ≥ 0 for every event, P(S) = 1, and for mutually exclusive events the probability of their union is the sum of their probabilities.

Classical probability

When outcomes are equally likely, P(E) is the number of favourable outcomes divided by the total number of outcomes in S.

Range of probability

Every probability lies between 0 and 1 inclusive; an impossible event has probability 0 and a certain event probability 1.

Addition theorem

For any two events, P(A∪B) = P(A) + P(B) − P(A∩B); the intersection is subtracted so it is not counted twice.

Mutually exclusive addition

If A and B cannot occur together then P(A∩B) = 0, so the addition theorem simplifies to P(A∪B) = P(A) + P(B).

Complement rule

Since A and A′ are exclusive and exhaustive, P(A) + P(A′) = 1, giving P(not A) = 1 − P(A), often the quickest route.