Probability Class 12 Mathematics Exam Questions

Please refer to Probability Class 12 Mathematics Exam Questions provided below. These questions and answers for Class 12 Mathematics have been designed based on the past trend of questions and important topics in your Class 12 Mathematics books. You should go through all Class 12 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 12 Mathematics Exam Questions Probability

Class 12 Mathematics students should read and understand the important questions and answers provided below for Probability which will help them to understand all important and difficult topics.

Very Short Answer Type Questions

1. A bag contains 10 white and 6 black balls. 4 balls are successively drawn out without replacement. What is the probability that they are alternately of different colours?
Answer. Required probability = P(BWBW) + P(WBWB)

2. A die is thrown repeatedly until a six comes up. Write the sample space for this experiment.
Answer. The sample space is
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 3, 6), ….}
It can be observed that infinite number of possibilities occur.

3. A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 yellow and 3 red balls. If it shows head, we throw a die. Write the sample space for this experiment.
Answer. Let H and T represent a head and a tail of a coin, respectively.
Also, let red balls are represented by R₁, R₂ and R₃ and
yellow balls are represented by Y₁ and Y₂. Then, the sample space,
S = {TR₁, TR₂, TR₃, TY1, TY2, H1, H2, H3, H4, H5, H6}

4. Find the total number of elementary events associated to the random experiment of throwing three die together.
Answer. When three dice are tossed together, then the total number of possible outcomes = 6³ = 6 × 6 × 6 = 216

5. A bag contains 4 identical red balls and 3 identical black balls. The experiment consists of drawing one ball, then putting it into the bag and again drawing a ball. Write the possible outcomes of this experiment.
Answer. The sample space for this experiment is
S ={RR, RB, BR, BB}, where R denotes the red ball and B denotes the black ball.

6. Two coins (a ₹ 2 coin and a ₹ 5 coin) are tossed once. Find the total number of elements in sample space.
Answer. The sample space is S = {HH, HT, TH, TT}
Thus, n(S) = 4

7. Let the sample space associated with an experiment is S = {1, 2, 3, 4, 5, 6} and an event is E = {1, 3, 5}, then find E′ or E.
Answer. Given that, S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}
∴ E = S – E = {2, 4, 6}.

8. Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.
Answer. The sample space S for selecting three bulbs at random from a lot is given by
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN }
where D indicates a defective bulb and N a non defective bulb.

9. The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.
Answer. The sample space S for the given experiment is
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}.

10. If A and B are two events associated with the same random experiment such that P(A ∪ B) = 3/4, P(A ∩ B) = 1/4 and P(A̅) = 2/3, then find P(B).
Answer.

Short Answer Type Questions -I

11. Three events A, B and C have probabilities

respectively. Given that P(A ∩ C) = 1/5 and P(B ∩ C) = 1/4 , find the value of P(C/B) and P(A̅∩C̅) .
Answer.

12. Two thirds of the students in a class are boys and the rest girls. It is known that the probability of a girl getting a first class is 0.25 and that of a boy getting a first class is 0.28. Find the probability that a student chosen at random will get first class marks in the subject.
Answer. Let E₁, E₂ and A be the events defined as follows:
E₁ = a boy is chosen from the class, E₂ = a girl is chosen from the class and, A = the student gets first class marks.
Then, P(E₁) = 2/3, P(E₂) = 1/3, P(A/E₁) = 0.28 and P(A/E₂) = 0.25.
Thus, P(A) = P(E₁) P(A/E₁) + P(E₂) P(A/E₂) = 0.27

13. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of respectively. Then, show that
(i) A is a simple event
(ii) B and C are compound events
(iii) A and B are mutually exclusive events.
Answer. We have, A = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(1, 3), (3, 1), (2, 2)}
(i) Since A consists of a single sample point, it is a simple event.
(ii) Since both B and C contain more than one sample point, therefore each one of them is a compound event.
(iii) Since A ∩ B = φ.
∴ A and B are mutually exclusive events.

14. Find the probability of getting the sum as a prime number when two dice are thrown together.
Answer. Total number of possible outcomes = 36
The sum should be prime number i.e., 2, 3, 5, 7, 11.
Sum 2 ≡ (1, 1)
Sum 3 ≡ (1, 2), (2, 1)
Sum 5 ≡ (1, 4), (4, 1), (2, 3), (3, 2)
Sum 7 ≡ (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)
Sum 11 ≡ (5, 6), (6, 5)
∴ Number of favourable outcomes = 15
Hence, required probability

15. A pair of dice is rolled. If the outcome is a doublet, a coin is tossed. Find the total number of possible outcomes for this experiment.
Answer. Corresponding to each doublet we get two outcomes.
For example, corresponding to (1, 1) we get (1, 1, H), (1, 1, T).
Thus, total number of possible outcomes = 36 + 6 = 42.

16. What is the probability that all L’s come together in the word PARALLEL?
Answer. Total number of possible outcomes

Total number of favourable outcomes when (3L’s consider as one letter) = 6!/2!
Hence, required probability

17. A card is drawn from a well shuffled deck of 52 cards, then find the probability of red king card.
Answer. The outcomes in the sample space S are 52.
∴ n(S) = 52
Let E be the event of getting a red king card
Out of 4 kings, 2 are red and 2 are black
∴ n(E) = 2
Hence, required probability

18. In a single throw of a die, find the probability of getting an even prime number.
Answer. Let S = {1, 2, 3, 4, 5, 6} be the sample space and E
= {2} be event of getting an even prime number.
⇒ n(S) = 6 & n(E) = 1
Hence, required probability

19. If A and B are events such that P(A) = 1/3, P(B) = 1/4 and P(A ∩ B) = 1/12, then find P(not A and not B).
Answer.

20. A university has to select an examiner from a list of 50 persons, 20 of them are women and 30 men, 10 of them knowing Hindi and 40 not, 15 of them being teachers and the remaining 35 not. What is the probability of the university selecting a Hindi knowing woman teacher?
Answer. Let E₁ : “A woman is selected”, E₂ : “A Hindi knowing person is selected” and E₃ : “A teacher is

Short Answer Type Questions -II

21. Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have non-adjacent desks?
Answer. Six employees can be seated in row in six desks in 6! ways. Married couple can occupy adjacent seats in the following 5 ways.
1 – 2, 2 – 3, 3 – 4, 4 – 5, 5 – 6
Also, they can interchange their seats and the remaining 4 seats can be occupied by remaining 4 employees in 4! ways.
∴ Number of ways in which married couple will have adjacent seats = 5 × 2! × 4!
So, number of ways in which married couple will have non-adjacent seats = 6! – 5 × 2! × 4! = 480

22. A coin is tossed three times, consider the following events :
A : ‘No head appears’
B : ‘Exactly one head appears’
C : ‘At least two heads appear’
Do they form a set of mutually exclusive and exhaustive events?
Answer. If S be the sample space of tossing a coin three times
then S = {HHH, HHT, HTH, THH, HTT, THT, TTH,TTT }
Now, A (Event of getting no head) = {TTT }
B (Event of getting exactly one head)
= {HTT, THT, TTH }
C (Event of getting atleast two heads)
= {HHT, HHH, HTH, THH }
Clearly, A ∩ B = f, B ∩ C = f and A ∩ C = f
So, A, B and C are mutually exclusive events.
Also, A ∪ B ∪ C = S
So, A, B and C are exhaustive events.

23. A bag contains 9 balls of which 4 are red, 3 are blue and 2 are yellow. The balls are similar in shape and size. A ball is drawn at random from the bag. Calculate the probability that it will be:
(i) red (ii) not blue
(iii) either red or blue
Answer. (i) Total number of possible outcomes = 9
Number of red balls = 4
∴ Number of favourable outcomes = 4
Hence, required probability = 4/9
(ii) Number of balls which are not blue = 4 + 2 = 6
∴ Number of favourable outcomes = 6
Hence, required probability = 6/9 = 2/3
(iii) Number of balls which are either red or blue
= 4 + 3 = 7
∴ Number of favourable outcomes = 7
Hence, required probability = 7/9

24. What is the probability that:
(i) a non-leap year has 53 Tuesdays?
(ii) a leap year has 53 Wednesdays?
(iii) a leap year has 53 Fridays and 53 Saturdays?
Answer. (i) Non leap year contains 365 days i.e.,
52 weeks + 1 day
52 weeks contain 52 Tuesdays.
53 Tuesdays means the remaining day is a Tuesday.
Total number of possibilities for remaining day = 7
Number of favourable outcomes = 1
∴ Probability for 53 Tuesdays = 1/7
(ii) Leap year contains 366 days = 52 weeks + 2 days
52 weeks contain 52 Wednesdays.
53 Wednesdays means that out of 2 remaining days one will be Wednesday.
Total number of possibilities for remaining two days
(SM, MT, TW, WTh, ThF, FSat, SatS) = 7
Number of favourable outcomes = 2
Hence, probability of 53 Wednesdays in a leap year = 2/7
(iii) Total possibilities for remaining two days of leap year (SM, MT, TW, WTh, ThF, FSat, SatS) = 7
Number of favourable outcomes = 1
Hence, required probability = 1

25. A shopkeeper sells three types of seeds A₁, A₂ and A₃. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of three types of seeds are 45%, 60% and 35%.
Calculate the probability
(i) that it will not germinate given that the seed is of type A3.
(ii) of a randomly chosen seed to germinate.
(iii) that it is of type A₂ given that a randomly chosen seed does not germinate.
Answer. Consider the following events:
E_i = Seed chosen is of type A_i, i = 1, 2, 3
A = Seed chosen germinates.

26. A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black.
Answer. Probability of choosing bag

27. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls? Given that
(i) the youngest is a girl.
(ii) atleast one is a girl.
Answer. Let G_i (i = 1, 2) and Bi (i = 1, 2) denote the i^th child is a girl or a boy respectively.
Then sample space is,
S = {G₁G₂, G₁B₂, B₁G₂, B₁B₂}
Let A be the event that both children are girls, B be the event that the youngest child is a girl and C be the event
that at least one of the children is a girl.
Then A = {G₁G₂}, B = {G₁G₂, B₁G₂}
and C = {B₁G₂, G₁G₂, B₁G₂}
⇒ A∩B = {G₁G₂} and A∩C = {G₁G₂}
(i) Required probability = P(A/B)

28. A couple has 2 children. Find the probability that both are boys, if it is known that
(i) atleast one of them is a boy,
(ii) the older child is a boy.
Answer. Let B_i(i = 1, 2) and G_i(i = 1, 2) denote the i^th child is a boy or a girl respectively.
Then sample space is,
S = {B₁B₂, B₁G₂, G₁B₂, G₁G₂}
Let A be the event that both are boys, B be the event that at least one of them is a boy and C be the event that the older child is a boy.
A = {B₁B₂}, B = {G₁B₂, B₁G₂, B₁B₂}
C = {B₁B₂, B₁G₂} ⇒ A∩B = {B₁B₂} and A∩C = {B₁B₂}
(i) Required probability = P(A/B)

29. A bag contains 3 red and 7 black balls. Two balls are selected at random one-by-one without replacement. If the second selected ball happens to be red, what is the probability that the first selected ball is also red?
Answer. Let A be the event of drawing a red ball in first draw and B be the event of drawing a red ball in second draw.

Now, P(B/A) = Probability of drawing a red ball in the second draw, when a red ball already has been drawn

30. P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are they likely to agree in stating the same fact?
Answer. Let E be the event that P speaks truth and F be the event that Q speaks truth. Then, E and F are independent

P and Q will agree to each other in stating the same fact in the following mutually exclusive ways:
(I) P speaks truth and Q speaks truth i.e. E ∩ F
(II) P tells a lie and Q tells a lie i.e. E̅ ∩ F̅.
∴ P(P and Q agree to each other)

Hence, in 62% of the cases P and Q are likely to agree in stating the same fact.

31. The probabilities of two students A and B coming to the school in time are 3/7 and 5/7 respectively. Assuming that the events, ‘A coming in time’ and ‘B coming in time’ are independent, find the probability of only one of them coming to the school in time.
Answer. Let E be the event that A is coming in time; P(E) = 3/7
and F be the event that B is coming in time, P(F) = 5/7
Also E and F are given to be independent events.
∴ Probability of only one of them coming to the
school in time = P(E) P(F̅) + P(E̅) P(F)

32. In a shop X, 30 tins of ghee of type A and 40 tins of ghee of type B which look alike, are kept for sale. While in shop Y, similar 50 tins of ghee of type A and 60 tins of ghee of type B are there. One tin of ghee is purchased from one of the randomly selected shop and is found to be of type B. Find the probability that it is purchased from shop Y.
Answer. Let E₁ be the event of getting ghee from shop X, E₂ be the event of getting ghee from shop Y and A be the event of getting ghee of type B.

33. Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4.
The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3 respectively. If the change does not take place, find the probability that it is due to the appointment of C.
Answer. Let I be the event that changes take place to improve profits.
Probability of selection of A, P(A) = 1/7
Probability of selection of B, P(B) = 2/7
Probability of selection of C, P(C) = 4/7
Probability that A does not introduce changes,
P(I̅ /A) = 1 − 0.8 = 0.2
Probability that B does not introduce changes,
P(I̅ /B) = 1 − 0.5 = 0.5
Probability that C does not introduce changes,
P(I̅ /C) = 1 − 0.3 = 0.7

34. Let X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that

where k is a positive constant. Find the value of k. Also find the probability that you will get admission in
(i) exactly one college
(ii) atmost 2 colleges
(iii) atleast 2 colleges.
Answer. The probability distribution of X is

(i) P (getting admission in exactly one college)
= P(X = 1) = k = 0.125
(ii) P (getting admission in atmost 2 colleges)
= P(X ≤ 2) = 0 + k + 4k = 5k = 0.625
(iii) P (getting admission in atleast 2 colleges)
= P(X ≥ 2) = 4k + 2k + k = 7k = 0.875

35. A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white ?
Answer. Consider the following events.
E : Two balls drawn are white
A : There are 2 white balls in the bag
B : There are 3 white balls in the bag
C : There are 4 white balls in the bag

Long Answer Type Questions

36. In a hockey match, both teams A and B scored same number of goals up to the end of the game, so as to decide the winner, the referee asked both the captains to throw a die alternately and decided that the team, whose captain gets a six first, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match.
Answer. Probability of getting a six by the captains of both the teams A and B is

Since A starts the game, he can throw a six in the following mutually exclusive ways :

37. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.
Answer. Let A, E₁ and E₂ respectively be the events that a person has a heart attack, the selected person followed the course of yoga and meditation and the person adopted the drug prescription

P(A/E₁) = 0.40 × 0.70 = 0.28,
P(A/E₂) = 0.40 × 0.75 = 0.30
Probability that the patient suffering from heart attack followed the course of meditation and yoga is

38. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain ‘A’ grade and 20% of day scholars attain ‘A’ grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an ‘A’ grade, what is the probability that the student is a hosteler?
Answer. Let E₁, E₂ and S be the following events :
E₁ : The student resides in hostel
E₂ : The student is a day-scholar
S : The student attains A grade.

39. Consider the experiment of tossing a coin. If the coin shows head, toss it again, but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.
Answer. The sample space S of the given random experiment is
S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
Let A be the event that the die shows a number greater than 4 and B be the event that there is at least one tail.
∴ A = {(T, 5), (T, 6)}
and B = {(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6),(H,T)}
A∩B ={(T, 5), (T, 6)}
∴ P(B) = P{(T, 1)} + P{(T, 2)} +P{(T, 3)}
+ P{(T, 4)} + P{(T, 5)} + P{(T, 6)} + P{(H, T)}

40. In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?
Answer. Let A, B, C and E are respectively the events that a person is smoker and non-vegetarian, smoker and vegetarian, non-smoker and vegetarian, and the selected person is suffering from the disease.
Here, n(A) = 160, n(B) = 100,
n(C) = 400 – (160 + 100) = 140.

CASE STUDY :

The reliability of a COVID PCR test is specified as follows:
Of people having COVID, 90% of the test detects the disease but 10% goes undetected. Of people free of COVID, 99% of the test is judged COVID negative but 1% are diagnosed as showing COVID positive. From a large population of which only 0.1% have COVID, one person is selected at random, given the COVID PCR test, and the pathologist reports him/her as COVID positive.
Based on the above information, answer the following

Question. What is the probability of the ‘person to be tested as COVID positive’ given that ‘he is actually having COVID?
a. 0.001
b. 0.1
c. 0.8
d. 0.9

Question. What is the probability of the ‘person to be tested as COVID positive’ given that ‘he is actually not having COVID’?
a. 0.01
b. 0.99
c. 0.1
d. 0.001

Question. What is the probability that the ‘person is actually not having COVID?
a. 0.998
b. 0.999
c. 0.001
d. 0.111

Question. What is the probability that the ‘person is actually having COVID given that ‘he is tested as COVID positive’?
a. 0.83
b. 0.0803
c. 0.083
d. 0.089

Question. What is the probability that the ‘person selected will be diagnosed as COVID positive’?
a. 0.1089
b. 0.01089
c. 0.0189
d. 0.189