Probability: Important Questions and Solved Examples

Probability is learned by solving, and the same question types appear in different disguises year after year. This guide collects the questions worth practising, groups them by type, and solves each one in full, with the reasoning shown, so that you can see how an answer is built and not only what it is.

If any formula here feels unfamiliar, read the full chapter notes on probability first. For further practice after this page, use the chapter MCQs and the important questions for the chapter, and for the next chapter’s ideas, the guide on mean, median and mode of grouped data.

How to Write a Probability Answer

A clean structure earns full marks and avoids small slips. Use the same four lines every time.

State the experiment and the total number of equally likely outcomes.
List or count the favourable outcomes.
Write the formula, P(E) = favourable ÷ total.
Substitute and simplify, and give the answer as a fraction in the lowest terms.

Type One: Single Die

Q1. A die is thrown once. Find the probability of getting (i) a prime number, (ii) a number between 2 and 6, (iii) an odd number.

Solution. Total outcomes = 6.

(i) The primes are 2, 3, 5, so there are 3. P = 3/6 = 1/2. (Remember that 1 is not prime.)
(ii) The numbers between 2 and 6 are 3, 4, 5, so there are 3. P = 3/6 = 1/2.
(iii) The odd numbers are 1, 3, 5. P = 3/6 = 1/2.

Type Two: Two Dice

With two dice, write the total as 6 × 6 = 36.

Q2. Two dice are thrown together. Find the probability that the sum is (i) a perfect square, (ii) greater than 9, (iii) a doublet.

Solution. Total = 36.

(i) The possible sums are 2 to 12, so the perfect squares are 4 and 9. A sum of 4 can be made in 3 ways, (1,3), (2,2), (3,1), and a sum of 9 in 4 ways, (3,6), (4,5), (5,4), (6,3). Favourable = 7. P = 7/36.
(ii) A sum greater than 9 means 10, 11 or 12, which can be made in 3 + 2 + 1 = 6 ways. P = 6/36 = 1/6.
(iii) A doublet means both dice show the same number: (1,1) to (6,6), which is 6 outcomes. P = 6/36 = 1/6.

Type Three: Coins

Q3. Three coins are tossed together. Find the probability of getting (i) exactly two heads, (ii) at most one head, (iii) at least two tails.

Solution. The outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, so the total is 8.

(i) Exactly two heads: HHT, HTH, THH. P = 3/8.
(ii) At most one head means zero or one head: TTT, HTT, THT, TTH. P = 4/8 = 1/2.
(iii) At least two tails is the same as at most one head. P = 1/2.

Type Four: Cards

Q4. One card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is (i) neither a red card nor a king, (ii) a red face card, (iii) the king or queen of hearts.

Solution. Total = 52.

(i) Red cards are 26, and the two black kings are not red, so “red or king” = 26 + 2 = 28. Neither = 52 − 28 = 24. P = 24/52 = 6/13.
(ii) Red face cards: 3 in hearts and 3 in diamonds, so 6. P = 6/52 = 3/26.
(iii) The king of hearts and the queen of hearts, which is 2 cards. P = 2/52 = 1/26.

The trick in (i) is not to double-count the red kings, which are already among the 26 red cards.

Type Five: Bags, Boxes and Lots

Q5. A bag contains 3 red, 5 white and 2 black balls. A ball is drawn at random. Find the probability that it is (i) white, (ii) not black, (iii) neither red nor white.

Solution. Total = 3 + 5 + 2 = 10.

(i) P(white) = 5/10 = 1/2.
(ii) Not black = 3 + 5 = 8. P = 8/10 = 4/5.
(iii) Neither red nor white means black. P = 2/10 = 1/5.

Q6. A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random. What is the probability that it is not defective?

Solution. Non-defective = 20 − 4 = 16. P = 16/20 = 4/5.

Q7. In a lottery, there are 10 prizes and 25 blanks. What is the probability of getting a prize?

Solution. Total tickets = 10 + 25 = 35. P = 10/35 = 2/7.

Type Six: Numbers and Letters

Q8. A number is chosen at random from 1 to 50. Find the probability that it is (i) a perfect square, (ii) a multiple of 7.

Solution. Total = 50.

(i) The perfect squares are 1, 4, 9, 16, 25, 36, 49, which is 7. P = 7/50.
(ii) The multiples of 7 are 7, 14, 21, 28, 35, 42, 49, which is 7. P = 7/50.

Q9. A letter is chosen at random from the word MATHEMATICS. Find the probability that it is (i) a vowel, (ii) the letter T.

Solution. The word has 11 letters: M, A, T, H, E, M, A, T, I, C, S.

(i) The vowels are A, E, A, I, which is 4. P = 4/11.
(ii) The letter T appears twice. P = 2/11.

Type Seven: The Calendar

Q10. What is the probability that a leap year selected at random has 53 Sundays?

Solution. A leap year has 366 days = 52 weeks + 2 extra days. The two extra days can be (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun) or (Sun, Mon), which gives 7 equally likely pairs. Two of these contain a Sunday. P = 2/7.

Type Eight: Algebra Meets Probability

These are the questions that surprise students, because the number of balls is an unknown.

Q11. A box contains 12 balls, of which x are black. If one ball is drawn at random, what is the probability that it is black? If 6 more black balls are put in the box, the probability of drawing a black ball becomes double the previous value. Find x.

Solution. The first probability is x/12. After adding 6 black balls, there are x + 6 black balls out of 12 + 6 = 18. The new probability is (x + 6)/18.

The condition says (x + 6)/18 = 2 × x/12 = x/6. Cross-multiplying: x + 6 = 3x, so 2x = 6 and x = 3.

Check. The first probability is 3/12 = 1/4. The new probability is 9/18 = 1/2, which is double 1/4. ✓

Q12. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls.

Solution. Let the number of blue balls be b, and the total be 5 + b. Then b/(5 + b) = 2 × 5/(5 + b). Multiplying both sides by (5 + b) gives b = 10. So there are 10 blue balls.

A Quick Reference of Methods

Question type	The method
“Not” or “neither… nor”	Count what is excluded, or use 1 − P(E)
“At least one”	Find P(none) and subtract from 1
“Or” with possible overlap	Count each outcome only once
Two dice	Total 36, and list the pairs
Cards	Use the facts: 52 cards, 26 red, 12 face cards, 4 of each kind
Unknown number of items	Write the probability as an expression and form an equation

Mistakes to Watch For

Taking two dice as 12 outcomes.
Treating 1 as a prime number.
Counting the red kings twice when finding “red or king”.
Forgetting that the probability after adding items changes both the favourable and the total.
Leaving fractions unsimplified.

Probability in Your Syllabus

Probability is the last chapter of the Class 10 Maths book and is retained in both the CBSE and Karnataka SSLC syllabi, according to published listings. It returns in Class 11 and Class 12, which feed into JEE Main and KCET, as in these Class 11 probability notes. Confirm the current portions with your own board.

Frequently Asked Questions

Q1. What are the important question types in Class 10 probability?

Single and double dice, coin tosses, cards, bags of coloured balls, numbers and letters, the leap year question, and algebra-based problems with an unknown number of items.

Q2. How do you find the probability of “at least one”?

Find the probability of “none” and subtract it from 1. For three coin tosses, P(at least one head) = 1 − 1/8 = 7/8.

Q3. What is the probability of getting a doublet when two dice are thrown?

There are 6 doublets out of 36 outcomes, so the probability is 6/36 = 1/6.

Q4. How many outcomes are there when two dice are thrown?

36, because each of the 6 outcomes of the first die pairs with each of the 6 outcomes of the second.

Q5. Is 1 a prime number in probability questions?

No. A prime number has exactly two factors, and 1 has only one, so it is not prime.

Practise the Pattern, Not the Question

The twelve questions above cover almost every variety the board paper uses. Solve each one again without looking, and for each one, say aloud which type it is. That recognition is worth more than any number of unseen questions.