# Probability Class 12 Mathematics Important Questions

Please refer to Probability Class 12 Mathematics Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Mathematics based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Mathematics for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.

## Class 12 Mathematics Important Questions Probability

Question. 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
So, required probability = P(C / I̅)

Question. 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 ?
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

Question. Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence find the mean of the distribution.
Let X denote the number of spade cards in a sample of 3 cards drawn from a well-shuffled pack of 52 cards.
Since there are 13 spade cards in the pack, so in a sample of 3 cards drawn, either there is no spade
card or one spade card or two spade cards or 3 spade cards. Thus X = 0, 1, 2 and 3.
Now, P(X = 0) = Probability of getting no spade card

Question. Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.
Answer. Let X be the random variable.
∴ X can take values 3, 4, 5, 6 or 7.
Total number of ways = 6C2 = 15
The probability distribution of a random variable X is given by

Question. In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII, what is the probability that a student chosen randomly studies in class XII, given that the chosen student is a girl?
Answer. Let E1 and E2 be the event that selected student is girl and other student respectively.

Let E be the event that selected student studies in class XII.

Question. In a game, a man wins ₹ 5 for getting a number greater than 4 and loses ₹ 1 otherwise, when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses.
Let X be the amount he wins/loses.
Then, X can take values –3, 3, 4, 5.
P(X = 5) = P(Getting a number greater than 4 in the

Question. 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) at most 2 colleges
(iii) at least 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 at most 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

Question. 12 cards, numbered 1 to 12 are placed in a box, mixed up thoroughly and then a card is drawn at random from the box. If it is known that the number on the drawn card is more than 3, then find the probability that it is an even number.
Answer. Sample space, S = {1, 2, 3, 4, ….., 12}
∴ n(S) = 12
Let A be the event that the number on the drawn card is more than 3.
∴ A = {4, 5, 6, 7, 8, 9, 10, 11, 12} and P(A) = 9/12
Also, let B be the event that the number on the drawn card is an even number.
∴ B = {2, 4, 6, 8, 10, 12}
Now, A ∩ B = {4, 6, 8, 10, 12} ∴ P(A ∩ B) = 5/12

Question. 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 in the first draw

Question. A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins.
Favourable outcomes for A to win
= {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

Question. Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that (i) The problem is solved (ii) exactly one of them solved the problem
Answer. Let X and Y denote the respective events of solving the given specific problem by A and B.
then P(X) = 1/2 and P (Y) = 1/3
(i) P (problem is solved)

Question. From a lot of 15 bulbs which include 5 defectives, a sample of 2 bulbs is drawn at random (without replacement). Find the probability distribution of the number of defective bulbs.
Let X denote the number of defective bulbs in a sample of 2 bulbs which are to be drawn.
Here number of defective bulbs = 5
Number of non-defective bulbs = 15 – 5 = 10
∴ X can take values 0, 1, 2.
Now, P(X = 0) = Probability of getting no defective bulb
= Probability of getting 2 non-defective bulbs.

Question. Three cards are drawn at random (without replacement) from a well shuffled pack of 52 playing cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.
Answer. Let X denote the number of red cards. So X can
take values 0, 1, 2, 3.
Total number of cards = 52
Number of red cards = 26.

Question. Three machines E1, E2 and E3 in a certain factory producing electric bulbs, produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the bulbs produced by each of machines E1 and E2 are defective and that 5% of those produced by machine E3 are defective. If one bulb is picked up at random from a day’s production, calculate the probability that it is defective.
Answer. Let A be the event that the bulb is defective.

Question. There are two bags, bag I and bag II. Bag I contains 4 white and 3 red balls while another bag II contains 3 white and 7 red balls. One ball is drawn at random from one of the bags and it is found to be white. Find the probability that it was drawn from bag I.
Let E1 and E2 be the events to select bag I and bag II respectively.

Question. A bag contains 7 red, 4 white and 5 black balls. Two balls are drawn at random and are found to be white. What is the probability that both the balls are white?
Answer. Total number of balls present in a bag
= (7 + 4 + 5) = 16
Probability of drawing two white balls is

Question. 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 with 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)}

Question. If A and B are two independent events such that P(A̅ ∩ B) = 2/15 and P(A ∩ B̅) = 1/6, then find P(A) and P(B).
Answer. It is given that A and B are independent events and P(A̅ ∩ B) = 2/15
⇒ P(A̅)P(B) = 2/15                                            ..(i)

Question. 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 and state whether the decision of the referee was fair or not.
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 :

Since the total probability is unity, the probability of team B winning the match

The decision of the referee was not fair as whosoever starts throwing the die gets an upper hand.

Question. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/5 be the probability that he knows the answer and 2/5 be the probability that he guesses. Assuming that a student who guesses the answer will be correct with probability 1/3, what is the probability that the student knows the answer given that he answered it correctly?
Answer. Let E1, E2 and A be the events defined as follows :
E1 : The student knows the answer
E2 : Th student guesses the answer
A : Th student answers correctly

Question. A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade.
Answer. Let E1, E2, E3, E4 and A be the events defined as below :
E1 : Missing card is a card of heart.
E2 : Missing card is a card of spade.
E3 : Missing card is a card of club.
E4 : Missing card is a card of diamond.
A : Drawing three spade cards from the remaining cards.

Question. There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows head. What is the probability that it was the two-headed coin?
Answer. Let A be the two-headed coin, B be the biased coin showing up heads 75% of the times and C be the biased coin showing up tails 40% (i.e., showing up heads 60%) of the times.
Let E1, E2 and E3 be the events of choosing coins of the type A, B and C respectively. Let S be the event of getting a head. Then

Question. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident for them are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver or a car driver?
Answer. Let the events are defined as
E1 : Person is a scooter driver
E2 : Person is a car driver
E3 : Person is a truck driver
A : Person meets with an accident.

Question. A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is ‘1’. Find the probability that it is actually 1.
Answer. Let E1 be the event that ‘1’ occurs, E2 be the event that ‘1’ does not occur and A be the event that man reports that it is ‘1’.

Now P(A/E1) be the probability that man reports that there is ‘1’ on the die and ‘1’ actually occurs.
= Probability that the man speaks the truth = 3/5 and P(A/E2) be the probability that man reports that there is ‘1’, when actually ‘1’ does not occur.
= Probability that man does not speak the truth

We have to find P(E1/A) i.e., probability that there is ‘1’ on the die given that the man has reported that there is ‘1’.

Question. 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. Interpret the result and state which of the above stated methods is more beneficial for the patient.
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

Since P(E1/A) < P(E2/A). So, the course of yoga and meditation is more beneficial for a person having heart attack.

Question. 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? What value is reflected in this question?
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.

The value reflected in this question is that being non-vegetarian and smokers has more chances of suffering from disease.

Question. 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.

Question. A girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin two times and notes the number of heads obtained. If she obtained exactly two heads, what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer. Let A1, A2 and A be the following events :
A1 : The girl gets 5 or 6 and hence tosses a coin 3-times.
A2 : The girl gets 1, 2, 3 or 4 and hence tosses a coin 2-times.
A : The girl gets exactly two heads.

Question. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be hearts. Find the probability of the missing card to be a heart.
Answer. Let E1, E2, E3, E4 be the events of the missing card to be of hearts, spade, club and diamond respectively and A be the event of drawing 2 heart cards from the remaining cards. Then

Question. Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male ? Assume that there are equal number of males and females.
Answer. Let E1 be the event of choosing a male, E2 be the event of choosing a female and A be the event that a person has grey hair.

Question. Bag I contains 3 red and 4 black balls and bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.
Answer. Let E1 be the event that bag I is chosen and E2 be the event that bag II is chosen and A be the event that the chosen ball is red.

Question. Given three identical boxes I, II and III each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
Answer. Consider the following events :
E1 : Box I is chosen
E2 : Box II is chosen
E3 : Box III is chosen
A : The coin drawn is of gold.
We have,

Question. There are three coins. One is a two-tailed coin (having tail on both faces), another is a biased coin that comes up heads 60% of the times and third is an unbiased coin. One of the three coins is chosen at random and tossed, and it shows tails. What is the probability that it was a two-tailed coin?
Answer. Let C1, C2 and C3 be the events of selecting a two-tailed coin, a biased coin and an unbiased coin.
Let T be the event that the coin appears with a tail.
Now, P(C1) = P(C2 ) = P(C3 ) = 1/3

Question. In a class, 5% of the boys and 10% of the girls have an IQ of more than 150. In the class, 60% of the students are boys and rest are girls. If a student is selected at random found to have an IQ of more than 150, then find the probability that the student is a boy.
E1 : Selected student is a boy
E2 : Selected student is a girl
A : e student has an IQ of more than 150

Question. Three bags contain balls as shown in the table below :

A bag is chosen at random and two balls are drawn from it. They happen to be white and red. What is the probability that they come from the III bag?
Answer. Let E1 be the event that bag I is chosen, E2 be the event that bag II is chosen, E3 be the event that bag III is chosen and A be the event that balls drawn are 1 white and 1 red ball.

Question. Five bad oranges are accidently mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.

Let X be the random variable denoting the number of bad oranges drawn.

Question. From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.
Answer. Total number of bulbs = 15
Number of defective bulbs = 5
P(defective bulb) = 5/15 = 1/3
Number of non-defective bulbs = 10
P(non-defective bulb) = 10/15 = 2/3
Let X be a random variable that denotes the number
of defective bulbs in a draw of 4 bulbs.
∴ X can take values 0, 1, 2, 3 and 4.
Now, using Binomial distribution,

Question. Five cards are drawn one by one, with replacement, from a well shuffled deck of 52 cards. Find the probability that
(i) all the five cards are diamonds,
(ii) only 3 cards are diamonds,
(iii) none is a diamond.
Total number of cards = 52,
Number of diamond cards = 13.
Let p denotes the probability of obtaining a diamond card in one trial.

Let X denote the number of diamond cards drwan.
Using Binomial Distribution, P(X = r) = nCr pr qn – r
(i) P(all the five cards are diamonds)

Question. From a lot of 10 bulbs which includes 3 defective, a sample of 2 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer. Let X denote the number of defective bulbs
drawn, then X can take values 0, 1, 2.
Probability of drawing defective bulb = 3/10
Probability of drawing non-defective bulb = 7/10

CASE STUDY :

A coach is training 3 players. He observes that the player A can hit a target 4 times in 5 shots, player B can hit 3 times in 4 shots and the player C can hit 2 times in 3 shots From this situation answer the following:

Question. Let the target is hit by A, B: the target is hit by B and, C: the target is hit by A and C. Then, the probability that A, B and, C all will hit, is
a. 4/5
b. 3/5
c. 2/5
d. 1/5

C

Question. Referring to (i), what is the probability that B, C will hit and A will lose?
a. 1/10
b. 3/10
c. 7/10
d. 4/10

A

Question. With reference to the events mentioned in (i), what is the probability that ‘any two of A, B and C will hit?
1. 1/30
2. 11/30
3. 17/30
4. 13/30

D

Question. What is the probability that ‘none of them will hit the target’?
a. 1/30
b. 1/60
c. 1/15
d. 2/15