Class 10 Maths · Original EZT practice
Probability (Class 10 Chapter 14): summary and practice
Probability assigns a number to chance: for equally likely outcomes, P(E) = favourable outcomes / total outcomes. The chapter fixes the landmarks - the impossible event at 0, the sure event at 1, everything else between - and the complement rule P(not E) = 1 - P(E), then applies them to coins, dice, cards and bags of balls. 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. The theoretical approach
When every outcome of an experiment is equally likely, P(E) = number of outcomes favourable to E / total number of outcomes. No experimenting needed - the count does the work. The experimental (empirical) idea sits inside the section's introduction as motivation.
2. The landmarks: 0 and 1
The impossible event has probability 0, the sure event has probability 1, and every event's probability lies between them. An answer outside [0, 1] is a miscount, not a surprise.
3. Complementary events
E and not E together cover every outcome, so P(not E) = 1 - P(E). When the direct count is long, counting the complement is often one line.
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 from a 52-card pack
Choose the method: P(E) = favourable / total; a standard pack has 52 cards, 4 suits, 13 ranks.
EZT practice question
A card is drawn from a well-shuffled pack of 52 cards. Find P(an ace).
Show the worked solution
- Total outcomes: 52, all equally likely after shuffling.
- Favourable: the 4 aces.
- P(ace) = 4/52.
Answer: 1/13.
Check: Each rank (ace through king) has 4 cards, so every rank carries the same 1/13 - the answer matches the pack's symmetry.
Watch for this: Well-shuffled is what makes the 52 outcomes equally likely - without that assumption the formula does not apply.
Now try it yourself
P(a king) from 52 cards?
Check your answer
1/13.
4 kings out of 52.
Balls from a bag
Choose the method: Count the balls of the wanted colour over all balls.
EZT practice question
A bag has 3 red and 5 blue balls. One ball is drawn. Find P(red).
Show the worked solution
- Total balls: 3 + 5 = 8.
- Favourable: 3 red.
- P(red) = 3/8.
Answer: 3/8 = 0.375.
Check: P(blue) = 5/8, and 3/8 + 5/8 = 1 - the two colours exhaust the bag, so the probabilities must sum to 1.
Watch for this: The denominator is ALL balls, not just the other colour - 3/5 would compare red to blue, not red to everything.
Now try it yourself
Same bag - P(blue)?
Check your answer
5/8.
5 blue out of 8.
Use the complement
Choose the method: P(not E) = 1 - P(E).
EZT practice question
In the 3-red-5-blue bag, find P(not red).
Show the worked solution
- P(red) = 3/8 from before.
- P(not red) = 1 - 3/8.
- = 5/8.
Answer: 5/8.
Check: Not red means blue here, and counting directly gives 5/8 - the complement rule matches the direct count.
Watch for this: The complement subtracts from 1, not from the total count - 8 - 3/8 mixes the two worlds.
Now try it yourself
P(not an ace) from 52 cards?
Check your answer
12/13.
1 - 1/13.
One fair coin
Choose the method: Two equally likely outcomes; each carries 1/2.
EZT practice question
A fair coin is tossed once. Find P(head) and P(tail).
Show the worked solution
- Outcomes: head, tail - equally likely for a fair coin.
- P(head) = 1/2.
- P(tail) = 1 - 1/2 = 1/2.
Answer: Both 1/2.
Check: Head and tail are complementary and exhaustive, so their probabilities must sum to 1 - they do.
Watch for this: Fairness is the assumption doing the work; a biased coin breaks the equal split.
Now try it yourself
A fair coin is tossed - P(not head)?
Check your answer
1/2.
Not head is tail.
A die: count the favourable faces
Choose the method: Six equally likely faces; count how many satisfy the event.
EZT practice question
A fair die is thrown. Find P(a number greater than 4).
Show the worked solution
- Outcomes: 1 to 6.
- Greater than 4: 5 and 6 - two outcomes.
- P = 2/6.
Answer: 1/3.
Check: P(at most 4) = 4/6 = 2/3, and 1/3 + 2/3 = 1 - complementary events balance.
Watch for this: Greater than 4 excludes 4 itself - read the inequality before counting.
Now try it yourself
P(a prime number) on a fair die?
Check your answer
1/2.
Primes 2, 3, 5 - 3/6.
Between 0 and 1, always
Choose the method: 0 means impossible, 1 means sure, and no probability can leave [0, 1].
EZT practice question
A bag holds only blue balls. What is P(red), and what is P(blue)?
Show the worked solution
- Red is impossible: no favourable outcomes, so 0/8 = 0.
- Blue is sure: every outcome favourable, so 8/8 = 1.
- Both sit exactly on the landmarks.
Answer: P(red) = 0, P(blue) = 1.
Check: 0 and 1 sum to 1 - the complement rule holds even at the extremes.
Watch for this: An answer bigger than 1 or below 0 always means a counting slip - probabilities cannot leave the interval.
Now try it yourself
P(drawing a 7 on a standard die)?
Check your answer
0.
No such outcome - the impossible event.
Try a mixed set
Choose a method or explanation before opening each answer.
1. P(E) for equally likely outcomes?
Show answer and reason
Favourable outcomes / total outcomes.
The theoretical definition.
2. P(not E)?
Show answer and reason
1 - P(E).
E and not E exhaust the outcomes.
3. P(sure event)?
Show answer and reason
1.
Every outcome is favourable.
4. Cards in a standard pack? Face cards?
Show answer and reason
52 cards; 12 face cards.
4 suits x 13 ranks; J, Q, K in each suit.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- P(a number greater than 4) on a fair die.
- P(a prime number) on a fair die.
- P(not an ace) from a 52-card pack.
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Questions about this practice page
- What does equally likely actually mean?
- Every outcome has the same chance - a fair coin's head and tail, a fair die's six faces, any card in a well-shuffled pack. The formula P(E) = favourable/total works only under this assumption; biased setups need a different approach.
- How does the complement rule save counting?
- When the event is long to count but its opposite is short, compute P(not E) and subtract from 1. P(not an ace) is one line from P(ace) = 1/13, while counting all non-aces directly takes 48 cards.
- Is this the same probability as in Class 11?
- The landmarks are shared, but the approaches differ. Class 10 uses the theoretical definition for equally likely outcomes directly; the Class 11 chapter builds the axiomatic approach - sample spaces, the algebra of events, and the general addition rule.
- Why can a probability never exceed 1?
- Favourable outcomes are a subset of all outcomes, so the fraction favourable/total can never top 1 - and it reaches 1 only when every outcome is favourable, the sure event.
- Are these the NCERT exercise questions for Probability (Class 10)?
- 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.