# Important Questions Class 11 Maths Chapter 16

## Important Questions for CBSE Class 11 Mathematics Chapter 16 – Probability

Probability is a measure of how likely an event is to occur. It is mathematically defined as the ratio of the number of favourable outcomes of an event to the total number of outcomes of the event.

The formula for calculating the probability of an event is to divide the “x” number of favourable outcomes by the “n” total number of possible outcomes. It can be represented as follows.

Event Probability = Favourable Outcomes/Total Outcomes = x/n

The number of favourable outcomes cannot exceed the total number of outcomes.

We frequently use probability in our daily lives without giving it much thought. Similarly, when it comes to using an informal method of probability, researchers take their time determining the accurate results of probability to assist healthcare providers, insurance companies, stakeholders, and many other critical decisions.

Probability can vary depending on how you approach the problem. When considering outcomes over a long period of repetition, the answer is 50:50. This demonstrates an important aspect of how probability works.

The topics that are covered in this chapter are as follows.

• Probability
• Introduction
• Random Experiments
• Event
• Axiomatic Approach to Probability

CBSE Class 11 Mathematics Chapter-16 Important Questions

Study Important Questions for Class 11 Mathematics Chapter 16 – Probability

Extramarks Important Questions for Class 11 Mathematics Chapter 16 is useful for students who want to study the chapter in a question-answer format. These questions are compiled by subject matter experts from NCERT books and past year’s question papers. These are written in a clear, concise manner so that the students understand how to write answers for examinations and learn the topic at hand.

Given below are Important Questions for 1, 4, and 6 marks each. To access the complete set of questions, click the link for the Extramarks website.

1 Mark

Q1. What are the odds that a leap year, chosen at random, will have 53 Sundays?

A1. The total number of days in a leap year is 366, with 52 complete weeks and two days extra. The two days could be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), or (Sunday, Monday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday, Tuesday).

P(53 Sundays in a leap year) = 2/7

Q2. Describe the test space if a coin is tossed, followed by a die.

A2. H1, H2, H3, H4, H5, T1, T2, T3, T4, T5, T6

T- Tails

Q3. We would like to select one child from a group of two boys and three girls. A coin is tossed, and if it comes up heads, a boy is selected; otherwise, a girl is selected. Describe the test space.

A3. HB1, HB2, TG1, TG2, TG3 are the answers.

TG- Tails Girl

Q4. A box contains one white ball and three identical black balls. Two balls are drawn at random, one after the other, with no replacement. Create the experiment’s sample space.

A4. S = {WB, BW, and BB.}

Where, WB- White Ball

BW- Black white

BB- Black Ball

4 Marks

Q1. Consider the following scenario: A coin is tossed three times. They form a set of mutually exclusive and exhaustive events if A: no head appears, B: exactly one head appears, and C: at least two heads appear.

A1. Let H represent the outcome of heads every time the coin is tossed.

Let T represent the outcome of tails every time the coin is tossed.

The sample size S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

From the given data, A = {TTT}, B={HTT, THT, TTH}, C = {HHT, HTH, THH, HHH}

A∪B∪C = S

Therefore, A, B, and C are exhaustive events.

Also, A∩B = ∅, A∩C = ∅, C∩C=∅, disjoint, i.e., they are mutually exclusive.

Q2. 40% of students in a school’s Class 11 study Mathematics, while 30% study Biology. 10% of the students in the class study both Mathematics and Biology. Determine whether a student will be studying Mathematics, Biology, or both subjects if selected randomly.

A2. Let P(M) represent the probability of students studying Mathematics.

Let P(B) represent the probability of students studying Biology.

P(M)  = 40100, P(B)  = 30100

P(M∩B)  = 10100

P(M∪B)  =  P(M)  +  P(B)  –  P(M∩B)

= 40100 + 30100 – 10100 =  60100= 0.6

Q3. A hockey game is played from 3 to 5 p.m. Write the problem if a man arrives late for the game and misses the only goal of the game, which is scored in the 20th minute?

A3. Total time = 3 pm – 5 pm =  2hr  =  120 min

He can see the goal only if he arrives within the initial 20 minutes.

P(he sees the goal) = 20/120=1/6

P(he does not see the goal) = 1−1/6=5/6

Q4. In a town of 6000 people, 1200 are over the age of 50, and 2000 are women. It is well known that 30% of females are over the age of 50. What is the likelihood that a randomly selected individual from the town is either female or over 30 years old?

A4. Let A1 be the event that the person is a female and A2 be the event that the person is 50 yr. old.

n(A1) = 2000,n(A2) = 1200

n(A1∩A2) = 30% of 2000 = 30/100×2000=600n(A1∪A2)  =  n(A1)  +  n(A2)  –  n(A1∩A2)

= 2000  +  120  –  600  =  2600

P(A1∪A2) = 2600/6000 = 1330

Q5. If an entrance exam is graded on the basis of two examinations, a randomly selected student has a 0.8 chance of passing the first examination and a 0.7 chance of passing the second examination. The chances of passing at least one of them are 0.95. What is the likelihood of passing both?

A5.  Let A represent the event where the student passes the first examination

Let B represent the event where the student passes the second examination

P(A) = 0.8,P(B) = 0.7

P(A∪B) = 0.95

P(A∩B) = ?

P(A∪B) = P(A) + P(B) – P(A∩B)

0.95 = 0.8 + 0.7 – P(A∩B)

0.55 = P(A∩B)

6 Marks

Q1. Three letters are dictated to three people, and each of them receives an envelope. The letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Determine whether or not at least one letter is in its proper envelope.

A1. Assume the three letters are A1, A2, and A3, and the three envelopes are E1, E2, and E3.

3P3 = 6 ways to put letters in three envelopes

Methods in which none of the letters is properly enveloped = 2

Probability:

P(at least one letter placed in an appropriate envelope) = 1 – P (none letters is put into a proper envelope)

= 1−2/6

= 2/3

Q2. The cards are numbered 1 to 20. One card is drawn at random. What is the probability that the number on the drawn card is

(i) a prime number; (ii) an odd number; (iii) a multiple of five; and (iv) not divisible by three?

A2. Let the sample space be ‘S’.

S = {1, 2, 3, 4,……., 20}

Let the events E1, E2, E3, and E4 each have a prime number, an odd number, a multiple of 5, and are not divisible by 3.

P(E1) = 820 = 25,E1 = { 2,3,5,7,11,13,17,19}

P(E2) = 1020 = 12,E2 = { 1,3,5,7,9,11,13,15,17,19}

P(E3) = 420 = 15,E3 = { 5,10,15,20}

P(E4) = 1420 = 710,E4 = { 1,2,4,5,7,8,10,11,13,14,16,17,19,20}

Q.1 Ashok and Amit play a game, where, each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is

Marks:1

Ans

The number of ways in which either of the players can choose from 1 to 25 is 25.
So, the total number of ways of choosing numbers = 25 25
= 625.
There are 25 ways in which the numbers chosen by both the players is the same.
Therefore, the probability that they will win a prize in a single trial = 25/625
= 1/25.
Hence, the probability that they will not win a prize in a single trial = 1 1/25
= 24/25.

Q.2 Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are kings, is

55/221.

105/663.

125/663.

135/663.

Marks:1

Ans

55/221.

Exhaustive no. of cases = 52C2 = 1326.
P(A) = Probability of red card
P(B) = Probability of king

Q.3 A, B and C are mutually exclusive and exhaustive events associated with a random experiment.
Find P(A) if P(B) = (5/2)P(A) and P(C) = (3/2)P(A).

Marks:1

Ans

Let P(A) = p
P(B) = (5/2) P(A)
= (5/2)P
and
P(C) = (3/2)P(A)
= (3/2)P
Since A, B and C are mutually exclusive and exhaustive events associated with a
random experiment, thus A ª B ª C = S
P( A ª B ª C) = P(S) = 1

P(A) + P(B) + P(C) = 1
p + (5/2)p + (3/2)p = 1
(10/2)p =1
p = 1/5

P(A) = p
= 1/5.

Q.4 A and B are two students. Probabilities of  solving a problem by A and B separately are  5/9 and 7/11 respectively. Find the probability that the problem will be solved.

Marks:4

Ans

Given that P(A) = 5/9 and (P(B) = 7/11
P(Ac) = 1 P(A)
= 4/9
P(Bc) = 1 P(B)
= 4/11
The problem will be solved if at least one of the two is able to solve the problem.

Therefore, Required Probability = P(A âˆª B)
= 1P(Ac). P(Bc)
= 1(4/9)— (4/11)
= 83/99

Q.5 There are 3 letters and 3 addressed envelopes. Find the probability that all the letters are not dispatched in the right envelopes.

Marks:4

Ans

3 letters can be put in 3 addressed envelopes in 3 ways.
3 letters can be put in 3 addressed envelopes correctly in 1 way.
Probability that all the letters are dispatched in the right envelopes = 1/3 = 1/6

Probability that all the letters are not dispatched in the right envelopes = 1 (1/6) = 5/6.