Previous year questions: Probability

Board pattern for this chapter

Probability is asked every year, usually as one or two short questions on conditional probability or independence and one longer question on distributions or Bayes' theorem. All ten items below are rewritten from official CBSE sources, six of them actual board questions from 2022, 2023 and 2024. Marks are printed in each heading, and the source path for each item is listed in report.md.

Q1 (CBSE Board 2022, 2 marks)

Two dice are thrown and the sum of the numbers on them is given to be even. Find the probability that the number on at least one die is 3.

Answer: There are 36 equally likely outcomes and 18 of them have an even sum. Among those, the cases with at least one 3 need the other die to be odd, giving (3,1), (3,3), (3,5), (1,3) and (5,3), five outcomes in all. The conditional probability is therefore 5/18.

Q2 (CBSE Board 2022, 2 marks)

Two people A and B try independently to solve a problem, with individual success probabilities 2/3 and 3/5. Find the probability that the problem is solved.

Answer: The problem is solved if at least one of them succeeds, so use the complement of both failing. The failure probabilities are 1/3 and 2/5, whose product is 2/15. The probability that the problem is solved is 1 - 2/15 = 13/15.

Q3 (CBSE Board 2022, 2 marks)

A basket has two rotten apples mixed with eight fresh ones. Two apples are drawn one after another without replacement. Find the probability distribution of the number of rotten apples drawn.

Answer: Let the random variable count the rotten apples among the two drawn. For no rotten apple both must be fresh: (8/10)(7/9) = 28/45. For two rotten apples both must be rotten: (2/10)(1/9) = 1/45. For exactly one rotten apple there are two orders, giving 2(2/10)(8/9) = 16/45. The three probabilities add to 1, as required.

Q4 (CBSE Board 2023, 3 marks)

Two independent events A and B are such that the chance of A without B is 1/4 and the chance of B without A is 1/6. Find the probabilities of A and B.

Answer: Write the two given conditions in terms of p, the probability of A, and q, the probability of B. Independence gives the overlap as pq, so p(1 - q) = 1/4 and q(1 - p) = 1/6. Subtracting gives p - q = 1/12, and substituting into the first equation gives a quadratic in q, namely 12q² - 11q + 2 = 0. The solutions are q = 2/3 or q = 1/4, with matching p = 3/4 or p = 1/3. Both pairs satisfy the given conditions and independence.

Q5 (CBSE Board 2024, 3 marks)

Two independent events E and F have the probability of not E equal to 0.6 and the probability of their union equal to 0.6. Find the probability of F and the probability that neither event occurs.

Answer: Since E has probability 0.4, independence makes the overlap 0.4 times the probability of F. The union therefore has probability 0.4 + P(F) - 0.4P(F) = 0.6, so 0.6P(F) = 0.2 and P(F) = 1/3. Neither event occurring has probability 0.6 times 2/3, which is 0.4.

Q6 (CBSE Board 2024, 4 marks)

Flights face severe, moderate or light turbulence with equal probability, and the chances of arriving late under these conditions are 55 per cent, 37 per cent and 17 per cent. (i) Find the probability that a flight arrives late. (ii) If a flight was late, find the probability that moderate turbulence was responsible.

Answer: The three turbulence levels form the partition, each with probability 1/3. (i) By total probability the answer is one third of the sum 0.55 + 0.37 + 0.17, which is 1.09/3, about 0.363. (ii) By Bayes' theorem the answer is one third of 0.37 divided by 1.09/3, which is 0.37/1.09, about 0.339.

Q7 (CBSE SQP 2021-22, 2 marks)

A bag holds one red and three white balls, and two balls are drawn one after another without replacement. Find the probability distribution of the number of red balls.

Answer: The random variable is 0 or 1. Both white has probability (3/4)(2/3) = 1/2. Exactly one red happens in either order, with probability (1/4)(1) + (3/4)(1/3) = 1/4 + 1/4 = 1/2. The distribution is therefore 1/2 for 0 and 1/2 for 1.

Q8 (CBSE SQP 2021-22, 2 marks)

Cards are drawn one by one from a well-shuffled pack of 52 without replacement. Find the probability that the first card is red and the second card is a jack.

Answer: Split the first draw into a red jack, of which there are 2, and a red card that is not a jack, of which there are 24. The probability is (2/52)(3/51) + (24/52)(4/51), which is 102/2652 = 1/26.

Q9 (CBSE SQP 2024-25, 5 marks)

A random variable takes all non-negative integral values, and the probability of the value r is proportional to one fifth raised to r. Find the constant of proportionality and the probability that the variable is less than 3.

Answer: Write the probability of r as k divided by 5 to the power r. Summing over r from 0 upwards gives k times the infinite geometric series 1 + 1/5 + 1/25 + ..., whose sum is 5/4, so k = 4/5. Then the probability of being less than 3 is k times (1 + 1/5 + 1/25) = (4/5)(31/25) = 124/125.

Q10 (CBSE Additional PQ 2021-22, 2 marks)

Out of 2000 applications under a genre and an age category, 960 were folk, 192 of the folk applications were below 18, and 104 of the classical applications were 18 and above. Given that an application is classical, find the probability that it is from the 18 and above category.

Answer: There are 1040 classical applications in all, and 104 of them are for the 18 and above category. The conditional probability is therefore 104/1040 = 1/10. The denominator is the classical total because the condition fixes the genre.

Worked model solution (three-case Bayes)

A workshop has three machines A, B and C producing 60, 30 and 10 per cent of the output, with defect rates 1, 2 and 5 per cent. An item chosen at random is defective. Find the probability that it came from machine C.

Let D be the event that the item is defective. The three machines partition the output, so the total probability of a defective item is 0.6(0.01) + 0.3(0.02) + 0.1(0.05) = 0.006 + 0.006 + 0.005 = 0.017. Bayes’ theorem gives P(C|D) = 0.005/0.017 = 5/17, about 0.294. The same denominator answers the other two posteriors: machine A gives 0.006/0.017 = 6/17 and machine B gives 6/17 as well. This is the shape of every Bayes answer in the paper, with the cases named, the prior line written, the likelihood line written, and one shared denominator.

What the board rewards

For conditional probability, name the conditioned event before substituting. For independence, show the product rule and, when the equation becomes quadratic, give both admissible pairs. For Bayes' theorem, list the cases, the priors and the conditionals, compute the shared total once, and then substitute. In distribution questions, start with the sum of probabilities equal to 1, then compute the mean, then the variance. A neat table and a stated formula name are worth marks even when the arithmetic is long.

Common mistakes

The most common mistake is reversing the condition, followed by assuming independence without checking the product. In Bayes questions, students often forget the denominator or use the wrong prior, and in distribution questions the frequent error is a table whose probabilities do not add to 1. In binomial questions, drawing without replacement is sometimes treated as Bernoulli trials; that is the single most serious mistake in the chapter because it changes the model completely.