Class 11 Maths · Original EZT practice
Probability (Class 11 Chapter 14): summary and practice
Probability formalises chance: random experiments and their sample spaces, events as subsets, the algebra of events, and the axiomatic rules - probabilities lie between 0 and 1, the sure event has probability 1, and mutually exclusive events add. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 22 October 2026:
1. Events and their types
An event is a subset of the sample space. Impossible (empty set), sure (whole space), simple (one outcome) and compound events - plus occurrence: an event occurs when the actual outcome belongs to it.
2. Algebra of events
Events combine like sets: A or B (union), A and B (intersection), not A (complement). Mutually exclusive events share no outcome; exhaustive events cover the whole sample space between them.
3. The axiomatic approach
Every outcome gets a probability between 0 and 1, all probabilities sum to 1, and probabilities of mutually exclusive events add. From these, everything else follows.
4. The working rules
For equally likely outcomes, P(E) = favourable/total. Then P(A or B) = P(A) + P(B) - P(A and B), and P(not A) = 1 - P(A).
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Probability of a simple event
Choose the method: P(E) = favourable outcomes / total outcomes, when outcomes are equally likely.
EZT practice question
A fair die is thrown. Find P(even number).
Show the worked solution
- Sample space: 1 to 6 - six equally likely outcomes.
- Even outcomes: 2, 4, 6 - three of them.
- P(even) = 3/6.
Answer: 1/2.
Check: Odd and even split the sample space evenly, so each must carry probability 1/2 - the answer matches the symmetry.
Watch for this: Count outcomes, not values: the die has 6 outcomes, and the event holds 3 of them.
Now try it yourself
P(multiple of 3) on a fair die?
Check your answer
1/3.
Outcomes 3 and 6 - 2/6.
Two coins: build the sample space first
Choose the method: List all ordered outcomes before counting.
EZT practice question
Two coins are tossed. Find P(exactly two heads).
Show the worked solution
- Sample space: HH, HT, TH, TT - four equally likely outcomes.
- Exactly two heads: HH only.
- P = 1/4.
Answer: 1/4.
Check: The four outcomes must sum to 1: HH, HT, TH, TT each carry 1/4, and exactly-two-heads is just one of them.
Watch for this: Two coins do NOT give three equally likely outcomes (two heads, one head, no heads) - HT and TH are distinct, so the space has 4 outcomes, not 3.
Now try it yourself
P(exactly one head) on two coins?
Check your answer
1/2.
HT and TH - 2/4.
Use the complement
Choose the method: P(not A) = 1 - P(A).
EZT practice question
If P(A) = 0.4, find P(not A).
Show the worked solution
- A and not A are mutually exclusive and exhaustive.
- Their probabilities sum to 1.
- P(not A) = 1 - 0.4.
Answer: 0.6.
Check: 0.4 + 0.6 = 1 - the pair must exhaust the sample space.
Watch for this: The complement is a subtraction from 1, not a reciprocal - 1/0.4 is meaningless here.
Now try it yourself
P(a king) = 1/13 from a 52-card pack. P(not a king)?
Check your answer
12/13.
1 - 1/13.
Add mutually exclusive events
Choose the method: If A and B cannot happen together, P(A or B) = P(A) + P(B).
EZT practice question
A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.5. Find P(A or B).
Show the worked solution
- Mutually exclusive means P(A and B) = 0.
- The addition rule applies directly.
- P(A or B) = 0.3 + 0.5.
Answer: 0.8.
Check: 0.8 stays below 1, so the pair does not overfill the sample space - consistent.
Watch for this: Check exclusivity BEFORE adding - if the events can overlap, plain addition double-counts the overlap.
Now try it yourself
Mutually exclusive C and D with P(C) = 0.2, P(D) = 0.7 - P(C or D)?
Check your answer
0.9.
0.2 + 0.7.
The general addition rule
Choose the method: P(A or B) = P(A) + P(B) - P(A and B).
EZT practice question
A card is drawn from a 52-card pack. Find P(a heart or a king).
Show the worked solution
- P(heart) = 13/52, P(king) = 4/52.
- The king of hearts is counted in both - P(heart and king) = 1/52.
- P(heart or king) = 13/52 + 4/52 - 1/52 = 16/52.
Answer: 16/52 = 4/13.
Check: Count directly: 13 hearts plus the 3 non-heart kings = 16 cards. The formula matches the direct count.
Watch for this: Forgetting the overlap counts the king of hearts twice - always subtract P(A and B).
Now try it yourself
P(a club or a queen) from 52 cards?
Check your answer
16/52 = 4/13.
Same structure: 13 + 4 - 1.
Mutually exclusive versus exhaustive
Choose the method: Exclusive: cannot happen together. Exhaustive: together they cover everything.
EZT practice question
In one throw of a die, let A = {1, 2} and B = {4, 5}. Are A and B mutually exclusive? Exhaustive?
Show the worked solution
- A and B share no outcome - mutually exclusive.
- Together they cover {1, 2, 4, 5}, leaving out 3 and 6.
- So not exhaustive.
Answer: Mutually exclusive, but not exhaustive.
Check: The two properties are independent: {1, 2} and {3, 4, 5, 6} would be both; {1, 2} and {2, 3, 4, 5, 6} would be exhaustive only.
Watch for this: Do not treat the words as synonyms - a pair can be one, both, or neither.
Now try it yourself
A = {1, 2}, B = {3, 4, 5, 6} on a die - exclusive? Exhaustive?
Check your answer
Both.
No overlap, and together they cover all six outcomes.
Try a mixed set
Choose a method or explanation before opening each answer.
1. P(not A)?
Show answer and reason
1 - P(A).
A and its complement exhaust the sample space.
2. P(A or B) in general?
Show answer and reason
P(A) + P(B) - P(A and B).
Subtract the double-counted overlap.
3. P(A or B) when mutually exclusive?
Show answer and reason
P(A) + P(B).
The overlap is zero.
4. Two coins - how many equally likely outcomes?
Show answer and reason
4 (HH, HT, TH, TT).
Order matters between the two coins.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- P(multiple of 3) on a fair die.
- P(exactly one head) when two coins are tossed.
- P(a club or a queen) from a 52-card pack.
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Questions about this practice page
- What is the difference between mutually exclusive and exhaustive events?
- Mutually exclusive events cannot happen together (no shared outcome); exhaustive events cover the whole sample space between them. A pair can be both - like even and odd on a die - or just one, like A = {1, 2} and B = {4, 5}, which are exclusive but leave out 3 and 6.
- Why must every probability lie between 0 and 1?
- The axioms fix it: an event is a subset of the sample space, so its probability is a share of the total probability 1. 0 marks the impossible event, 1 the sure event, and everything real sits between.
- How do I avoid sample-space mistakes with two coins or two dice?
- List the ordered outcomes explicitly. Two coins give HH, HT, TH, TT - four outcomes, not three, because HT and TH are distinct. Two dice give 36 ordered pairs. Count first, then assign probabilities.
- When can I just add probabilities?
- Only when the events are mutually exclusive - when they cannot happen together. Otherwise use P(A or B) = P(A) + P(B) - P(A and B) so the overlap is not counted twice.
- Are these the NCERT exercise questions for Probability (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.