Previous year questions: Probability

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each one uses the ideas from the first page, and every answer is worked out step by step. The set covers counting favourable outcomes, the axioms, complementary events and the addition rule.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The numbers a, b and c are chosen independently from {1, 2, 3, 4}. Find the probability that the quadratic ax squared plus 2 root 2 b x plus c is positive for every real x. If this probability is m/n in lowest terms, find m + n.

Answer: There are 4 cubed, that is 64, equally likely ordered triples. Since a is positive, the quadratic is positive for every real x exactly when its discriminant is negative: 8 b squared minus 4ac less than 0, that is ac greater than 2 b squared. For b = 1 the condition ac greater than 2 removes the three pairs (1,1), (1,2) and (2,1) from 16, giving 13 pairs. For b = 2 the condition ac greater than 8 gives 4 pairs, and for b = 3 and b = 4 no pair satisfies ac greater than 18. The favourable triples number 13 + 4 = 17, so p = 17/64 and m + n = 81. The answer is 81. The count for b = 2 is four because the pairs with product greater than 8 are (3,3), (3,4), (4,3) and (4,4).

Q2 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A fair coin is tossed repeatedly. Each head gives 10 points and each tail gives 5 points. Find the probability that the score becomes exactly 30 at some stage. If this probability is m/n in lowest terms, find m + n.

Answer: Let P(s) be the probability of eventually reaching exactly 30 from a score s below 30, with the rule that a score above 30 can never return. Then P(30) = 1 and P(s) = half of P(s + 10) plus half of P(s + 5). Working backwards: P(25) = 1/2, P(20) = 3/4, P(15) = 5/8, P(10) = 11/16, P(5) = 21/32 and P(0) = 43/64. Hence m + n = 43 + 64 = 107. The recursion works because after each toss the score moves to one of two values and the future probability depends only on the current score.

Q3 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

A coin is tossed eight times. Let p be the probability that exactly four heads appear in the first six tosses and exactly three heads appear in the last five tosses. Find 96p.

Answer: Let x be the number of heads among tosses 4, 5 and 6, which are common to the two groups. Then the first three tosses contain 4 - x heads and the last two contain 3 - x heads. The values x = 1, 2 and 3 are possible. Counting sequences, the favourable count is 3 + 18 + 3 = 24 out of 2 to the power 8, which is 256, so p = 24/256 = 3/32. Therefore 96p = 9. The three cases x = 1, 2, 3 correspond to the numbers of heads in the shared middle tosses, and each case chooses positions independently in the three groups of tosses.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

A bag contains N + 1 coins: N fair coins and one coin with heads on both faces. A coin is picked at random and tossed. The probability of getting a head is 9/16. Find N.

Answer: A fair coin gives a head with probability 1/2 and the double-headed coin always gives a head. Weighting by the chance of picking each coin: (N/(N+1))(1/2) + (1/(N+1))(1) = 9/16. The left side simplifies to (N + 2)/(2(N + 1)), so 16(N + 2) = 18(N + 1), which gives 16N + 32 = 18N + 18 and N = 7. The equation is linear, so there is exactly one solution, and checking N = 7 in the original probability gives 9/16 as required.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

A bag holds 6 blue and 6 green balls. Balls are drawn in pairs without replacement until the bag is empty. Find the probability that every drawn pair contains one blue and one green ball.

Answer: For the first pair the probability of one blue and one green is 36/66 = 6/11. After removing one of each colour, the bag holds 5 blue and 5 green, so the next probability is 25/45 = 5/9. The successive factors are 6/11, 5/9, 4/7, 3/5, 2/3 and 1. Multiplying and cancelling gives 16/231. The correct option is 16/231. The product telescopes because each factor after the first removes one ball of each colour, so the numerator and denominator share consecutive factors.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

A candidate travels to an examination centre by bus, scooter or car with probabilities 2/5, 1/5 and 2/5. The probabilities of arriving late by these three means are 1/5, 1/3 and 1/4. Given that the candidate arrived late, find the probability that the bus was used.

Answer: The probability of arriving late is (2/5)(1/5) + (1/5)(1/3) + (2/5)(1/4) = 2/25 + 1/15 + 1/10. Taking a common denominator of 150 gives 12/150 + 10/150 + 15/150 = 37/150. The probability of late by bus is 2/25 = 12/150. Dividing gives (12/150)/(37/150) = 12/37. The correct option is 12/37.

Q7 (NCERT Exemplar, 3 marks)

A family has exactly three children, each equally likely to be a boy or a girl. Find the probability that exactly one child is a girl, that at least two are girls, and that none is a girl.

Answer: The sample space is {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}, eight equally likely outcomes. Exactly one girl occurs in BBG, BGB and GBB, so the probability is 3/8. At least two girls occurs in BGG, GBG, GGB and GGG, so the probability is 4/8 = 1/2. No girl occurs only in BBB, so the probability is 1/8. The three probabilities 3/8, 1/2 and 1/8 correspond to finishing the listing of the eight outcomes without repetition.

Q8 (NCERT Exemplar, 2 marks)

How many two-digit positive integers are multiples of 3, and what is the probability that a randomly chosen two-digit positive integer is a multiple of 3?

Answer: The two-digit multiples of 3 run from 12 to 99 with common difference 3. The count is (99 - 12)/3 + 1 = 30. The two-digit positive integers run from 10 to 99, a total of 90. Since each integer is equally likely, the probability is 30/90 = 1/3. The count 30 follows from the arithmetic progression 12, 15, ..., 99, which has 30 terms because (99 - 12)/3 + 1 = 30.

Q9 (NCERT Exemplar, 2 marks)

At a road block the probability that a truck has faulty brakes is 0.23, the probability of badly worn tyres is 0.24, and the probability of at least one of the two problems is 0.38. Find the probability that a truck has both problems.

Answer: Write B for faulty brakes and T for worn tyres. The addition rule gives P(B union T) = P(B) + P(T) - P(B intersection T), so 0.38 = 0.23 + 0.24 - x. Hence x = 0.09, and nine percent of the trucks have both problems. The probability of neither problem is 1 - 0.38 = 0.62.

Q10 (NCERT Exemplar, 1 mark)

In a leap year, find the probability of having 53 Sundays or 53 Mondays.

Answer: A leap year has 366 days, that is 52 complete weeks plus 2 extra days. The two extra days can be Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday or Saturday-Sunday, seven equally likely pairs. The year has 53 Sundays or 53 Mondays exactly when the extra pair contains Sunday or Monday, which happens for Sunday-Monday, Monday-Tuesday and Saturday-Sunday, three pairs. The probability is 3/7. There are seven equally likely pairs of extra days, and three of them contain Sunday or Monday, so the favourable count is three.