CBSE Maths › Class 12 PYQs › Probability

Probability: 4 marks Questions (CBSE Class 12)

5 different 4 marks questions on Probability from CBSE Class 12 Maths board exams 2026, newest first.

Q374 marksCase StudyProbability CBSE 2026 · 65/1/1

In an online jackpot, there is one first prize of ₹ 3,00,000, two second prizes of ₹ 2,00,000 each and three third prizes of ₹ 50,000 each.
A total of 1,00,000 jackpot tickets each costing ₹ 100 were sold there by raising a fund of ₹ 1,00,00,000.
Rohan bought one ticket.
Based on given information, answer the following questions :
(i) What are the possible amounts, the person can win ? (1)
(ii) What is the probability that the person wins atleast ₹ 2,00,000 ? (2)
OR (ii) What is the probability that the person does not win any amount ? (2)
(iii) In another jackpot, Rohan also bought a ticket having a prize money of ₹ 5,00,000. The chances of winning the jackpot are 1 in 1,00,000. Find the probability that on exactly one of tickets he wins the jackpot. (1)

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Answer: (i) ₹ 3,00,000, ₹ 2,00,000, ₹ 50,000 (or ₹ 0) (ii)

\frac{3}{100000}

; OR

\frac{99994}{100000} = \frac{49997}{50000}

(iii)

\frac{2 \times 99999}{1 0 ^{10}} = \frac{99999}{5 \times 1 0 ^{9}}

(i) He can win ₹ 3,00,000, ₹ 2,00,000 or ₹ 50,000, or nothing (₹ 0).
(ii) Tickets winning at least ₹ 2,00,000: 1 + 2 = 3. Probability $= \frac{3}{100000}$ .
OR (ii) Winning tickets: 1 + 2 + 3 = 6. P(no prize) $= \frac{100000 - 6}{100000} = \frac{49997}{50000}$ .
(iii) Take P(jackpot on the first ticket) = P(first prize) $= \frac{1}{100000}$ and P(jackpot on the second ticket) $= \frac{1}{100000}$ , independent.
P(exactly one) $= 2 \times \frac{1}{100000} \times \frac{99999}{100000} = \frac{199998}{1 0 ^{10}} = \frac{99999}{5 \times 1 0 ^{9}}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q364 marksCase StudyProbability CBSE 2026 · 65/2/1

Smoking increases the risk of lung problems.
A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males.
A person is selected at random from these 50 people and tested for lung related problems.
Based on the given information, answer the following questions :
(i) What is the probability that selected person is a female ? (1)
(ii) If a male person is selected, what is the probability that he will not be suffering from lung problems ? (1)
(iii) (a) A person selected at random is detected with lung complications. Find the probability that selected person is a female. (2)
OR (iii) (b) A person selected at random is not having lung problems, find the probability that the person is a male. (2)

Show answer & solution

Answer: (i)

\frac{2}{5}

(ii)

\frac{83}{100}

(iii)(a)

\frac{8}{25}

OR (iii)(b)

\frac{249}{425}

Let M, F: male, female; L: lung problems. $P (M) = \frac{30}{50} = \frac{3}{5}$ , $P (F) = \frac{2}{5}$ , $P (L ∣ M) = 0.17$ , $P (L ∣ F) = 0.12$ .
(i) $P (F) = \frac{20}{50} = \frac{2}{5}$ .
(ii) $P (L^{'} ∣ M) = 1 - 0.17 = 0.83 = \frac{83}{100}$ .
(iii)(a) $P (L) = \frac{3}{5} (0.17) + \frac{2}{5} (0.12) = 0.102 + 0.048 = 0.15$ ; $P (F ∣ L) = \frac{0.048}{0.15} = \frac{8}{25}$ .
(iii)(b) $P (L^{'}) = \frac{3}{5} (0.83) + \frac{2}{5} (0.88) = 0.498 + 0.352 = 0.85$ ; $P (M ∣ L^{'}) = \frac{0.498}{0.85} = \frac{249}{425}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q374 marksCase StudyProbability CBSE 2026 · 65/3/1

A survey was conducted to find out the success rate of students who qualified the entrance examination by dropping a year after class XII.
As per the data collected, 40% students appearing in the examination were dropouts and the remaining students were regular students of class XII.
Of the dropouts, 5% qualify the examination while 10% of the regular students qualify the examination.
Based on the above information, answer the following questions.
(i) Find the probability that a student selected at random is a regular student. (1)
(ii) A student is selected at random from a group of dropout students. What is the probability that the student will not qualify the examination ? (1)
(iii) (a) A student selected at random qualified the examination. Find the probability that student is not a dropout. (2)
OR (iii) (b) A student selected at random did not qualify the examination. Find the probability that the student was a regular student. (2)

Show answer & solution

Answer: (i) 0.6 (ii) 0.95 (iii)(a)

\frac{3}{4}

(iii)(b)

\frac{27}{46}

Let D = dropout, R = regular, Q = qualifies. $P (D) = 0.4$ , $P (R) = 0.6$ , $P (Q ∣ D) = 0.05$ , $P (Q ∣ R) = 0.1$ .
(i) $P (R) = 0.6$ .
(ii) $P (Q^{'} ∣ D) = 1 - 0.05 = 0.95$ .
(iii)(a) $P (Q) = 0.4 (0.05) + 0.6 (0.1) = 0.02 + 0.06 = 0.08$ ; $P (R ∣ Q) = \frac{0.06}{0.08} = \frac{3}{4}$ .
(iii)(b) $P (Q^{'}) = 0.4 (0.95) + 0.6 (0.9) = 0.38 + 0.54 = 0.92$ ; $P (R ∣ Q^{'}) = \frac{0.54}{0.92} = \frac{27}{46}$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q384 marksCase StudyProbability CBSE 2026 · 65/4/1

An NGO organises a charity event in which they decide to distribute woollen caps to protect children from winter. The caps to be distributed are in three separate boxes, Box I has 30 red caps, Box II has 20 red and 10 green caps, and Box III has 30 green caps. The probability that a Box i is selected and a cap picked out is $\frac{i}{6}$ , where i = 1, 2, 3.
Based on the above information, answer the following questions :
A person selects a cap.
(i) What is the probability that he selects a red cap ? (2)
(ii) If he selects a green cap, what is the probability that the cap has come from Box II ? (2)

Show answer & solution

Answer: (i)

\frac{7}{18}

(ii)

\frac{2}{11}

Let $E_{i}$ : Box i is selected; $P (E_{1}) = \frac{1}{6}$ , $P (E_{2}) = \frac{2}{6}$ , $P (E_{3}) = \frac{3}{6}$ .
(i) $P (R ∣ E_{1}) = 1$ , $P (R ∣ E_{2}) = \frac{2}{3}$ , $P (R ∣ E_{3}) = 0$ .
$P (R) = \frac{1}{6} (1) + \frac{2}{6} \cdot \frac{2}{3} + \frac{3}{6} (0) = \frac{3}{18} + \frac{4}{18} = \frac{7}{18}$ .
(ii) $P (G) = 1 - \frac{7}{18} = \frac{11}{18}$ ; $P (G ∣ E_{2}) = \frac{1}{3}$ .
$P (E_{2} ∣ G) = \frac{\frac{2}{6} \cdot \frac{1}{3}}{\frac{11}{18}} = \frac{\frac{1}{9}}{\frac{11}{18}} = \frac{2}{11}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q374 marksCase StudyProbability CBSE 2026 · 65/5/1

There are three types of vaccines $A_{1}$ , $A_{2}$ , $A_{3}$ , available in the market to protect the population of the country from spread of certain infection. According to a survey conducted, it was found that 25% of the population was given Vaccine $A_{1}$ , 35% of the population was given Vaccine $A_{2}$ and 40% of the population was given Vaccine $A_{3}$ . The survey also stated that the probabilities that Vaccines $A_{1}$ , $A_{2}$ and $A_{3}$ would protect against the infection were 60%, 55% and 50% respectively.
Based on the above information, answer the following questions :
Find the probability that :
(i) The person taking vaccine $A_{2}$ will get infected. (1)
(ii) If a person is chosen randomly, he/she will be protected from the infection. (1)
(iii) (a) The person was given Vaccine $A_{1}$ , given that the randomly chosen person is infected. (2)
OR (iii) (b) The person was given Vaccine $A_{3}$ , given that the randomly chosen person is not infected. (2)

Show answer & solution

Answer: (i) 0.45 (ii) 0.5425 (iii) (a)

\frac{40}{183}

OR (iii) (b)

\frac{80}{217}

$P (A_{1}) = 0.25$ , $P (A_{2}) = 0.35$ , $P (A_{3}) = 0.40$ ; P(protected | $A_{i}$ ) = 0.60, 0.55, 0.50.
(i) P(infected | $A_{2}$ ) = 1 − 0.55 = 0.45.
(ii) P(protected) = 0.25(0.60) + 0.35(0.55) + 0.40(0.50) = 0.15 + 0.1925 + 0.2 = 0.5425.
(iii) (a) P(infected) = 1 − 0.5425 = 0.4575; P( $A_{1}$ | infected) = $\frac{0.25 \times 0.40}{0.4575} = \frac{0.1}{0.4575} = \frac{40}{183}$ .
(iii) (b) P( $A_{3}$ | not infected) = $\frac{0.40 \times 0.50}{0.5425} = \frac{0.2}{0.5425} = \frac{80}{217}$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

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