Class 11 Probability

What are the chances that India will win the match today? Do you think it will rain today? Answering these questions can be tricky as they depend upon possible outcomes that are likely to occur for a particular event. Class 11 Probability will familiarize you with concepts and formulas that can help you with these predictions. The chapter covers a range of examples to help students understand the concept of Probability. This blog covers Probability study notes for class 11 along with some solved examples.

Class 11 Probability: Overview

The term probability is defined as a continuous measure of various uncertain events. It can be separated by using two prominent theories: the classical theory and the statistical approach of probability. As per class 11 maths syllabus, students can also explore probability in contrast with the axiomatic approach with suitable examples. The ultimate concept of class 11 chapter on probability is based on the formula:

Probability= Possible or average number of outcomes/ Total number of outcomes.

Class 11 Probability: Probability Conditions

These probability conditions are derived from the probability chapter and are really elaborative in nature so that you can easily prepare for the exams or while solving problems in general.

Condition: Probability Conditions: A Coin is Tossed Thrice

Solution:
A coin has 2 faces, one head (H) and another tail (T)
When the coin is tossed three times the actual number of its possible outcomes are 2×3=6 
So, when the coin is tossed thrice, it’s sample space is given as-
S =( HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)

Condition: A Coin is Tossed, and a Die is Thrown

Solution:
A coin has 2 faces, one head (H) and another tail (T)
A die has a total of 6 faces numbering from 1 to 6, and one number faces each face
So, when the coin is tossed, and a die is thrown the sample space is given as-
S =( H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6)

Condition: If the Dice is Thrown Twice

Solution:
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6
When a die is thrown twice its sample space is given by:
S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
The total no. of elements in this sample space => 6 × 6 = 36, while the sample space is given by:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3),(3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Condition: A Coin is Tossed Four Times

Solution:
Since either coin can turn up, Head (H) or Tail (T) are the possible outcomes. When 1 coin is tossed one time, its sample space is 2
Then, when the coin is tossed 3 times, the sample space is 24 = 16
Thus, the sample space,
S = {HHHH, THHH, HTHH, HHTH, HHHT, TTTT, HTTT, THTT, TTHT, TTTH, TTHH, HHTT, THTH, HTHT, THHT, HTTH}

Condition: Various Colours Dice
One dice of red colour, one of white colour, one of blue colour are placed in a bag. One die is selected at random and rolled, it’s colour, and the number on its uppermost face is noted. Discuss this sample space.

Solution:
Let us assume that 1, 2, 3, 4, 5, 6 are the possible numbers that come when the die is thrown
As per the class 11 chapter on probability, assume die of red colour be ‘R,’ die of white colour be ‘W,’ die of blue colour be ‘B.’ Hence, the total no. of sample space will be 6 x 3 = 18
The sample space of the event is
S={(R,1),(R,2),(R,3),(R,4),(R,5),(R,6),(W,1),(W,2),(W,3),(W,4),(W,5),(W,6) (B,1),(B,2),(B,3),(B,4),(B,5),(B,6)}

Note: ‘Die’ is singular in form whereas ‘Dice’ is plural in nature.

Class 11 Straight Lines

Class 11 Probability: Conditional Probability

The class 11 chapter on probability states that conditional probability and mutually exclusive events must be given due consideration from the examination point of view. The conditional probability represents the probability of an event taking place in a relationship to one or multiple events. For example, A and B are the two consecutive events associated with a similar sample space of any random experiment. The conditional probability for the event D will be under the condition where E has offered will be written as P (D | E)

Example: A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

Solution: Let us assume that 1, 2, 3, 4, 5, and 6 are the possible outcomes when the die is thrown.
As per the condition mentioned in the question, if a die is thrown repeatedly until six appears.
If six may come up for the first throw or six may come upon the second throw, this process will go continuously until the six comes.

The SS (sample space) when 6 appears on very first throw S1 = {6}
The SS (sample space) when 6 appears on second throw S2 = {(1,6), (2,6), (3,6), (4,6), (5,6)}
This event can go on for infinite times
So, the sample space is infinitely defined
S = {(6), (1,6), (2,6), (3,6), (4,6), (5,6), (1,1,6), (1,2,6)……}

Permutation and Combination Class 11

What Are Conditional Probability Independent Events?

For any two events, A and B associated with a sample space D. If the probability of the random occurrence of any one of the events is not affected by the other event’s occurrence, both of them are referred to as independent events. On the contrary, mutually exclusive events are the opposite representation of the same. A random variable refers to the function of a real-valued variable. The domain of this variable is the actual sample space of any random experiment.

Example: Let E and F be events with P (E) = 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5. Are E and F independent?

Solution: As per the class 11 maths NCERT solutions for this exercise-
P (E)= 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ≠ 1/5  
⇒ P (E ∩ F) ≠ P (E). P (F)

Class 11 Probability Sample Questions

1. If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

Ans: We have word ALGORITHM Number of letters = 9

2. A die is rolled in such a way that each odd number is twice as likely to occur as each even number.

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Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Ans: If is given that 2 x Probability of even number = Probability of odd number

3. Four candidates A, B, C, ZJhave applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that
(a) C will be selected? (b) A will not be selected?

Ans: It is given that A is twice as likely to be selected as B
P(A) = 2P(B)
B and C are given about the same chance of being selected
P(B) = P(C)
C is twice as likely to be selected as D
P(C) = 2 P(D)

4. Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

5. While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours

6. If 6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is

7. A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is

8. If the probabilities for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is

9. The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get a compartment is 0.96.

Answer: False
Let A = Student will pass examination
B = Student will getting compartment
P(A) = 0.73, P(B) = 0.13 and P(A or B) = 0.96
P(A or B) = P(A) + P(B) = 0.73 + 0.13 = 0.86
But P(A or B) = 0.96
Hence, the given statement is false.

10. The probability of intersection of two events A and B is always less than or equal to those favourable to the event

Answer: True
We know that A ∩ B ⊂ A
P(A ∩ B) ≤ P(A)
Hence, it is a true statement.

11. The probability of an occurrence of event A is .7 and that of the occurrence of event B is .3 and the probability of occurrence of both is .4.

Answer: False
A ∩B⊆ A, B
P(A ∩B) ≤ P(A), P(B)
But given that P(B) = 0.3 and P(A ∩B) = 0.4, which is not possible.

12. The sum of probabilities of two students getting distinction in their final examinations is 1.2.

Answer: True
Probability of each student getting distinction in their final examination is less than or equal to 1, the sum of the probabilities of two may be 1.2.
Hence, it is a true statement.

13. A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

Answer: In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2, 5, 6), … ,(5, 1, 6), (5, 2, 6), …}

14. A coin is tossed twice, what is the probability that at least one tail occurs?

Answer: When a coin is tossed twice, the sample space is given by
S = {HH, HT, TH, TT}
Let A be the event of the occurrence of at least one tail
Accordingly, A = {HT, TH, TT}

15. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

Answer: Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Accordingly, n(S) = 12
(i) Let A be the event in which the sum of numbers that turn up is 3
Accordingly, A = {(1, 2)}

(ii) Let B be the event in which the sum of numbers that turn up is 12
Accordingly, B = {(6, 6)}

Thus, we hope that after going through this blog on probability for class 11, you are now familiar with the topic. Our career experts at Leverage Edu are here to guide you making the right career choices in your dream field. Hurry up! Book an e-meeting with us!

What are the chances that India will win the match today? Do you think it will rain today? Answering these questions can be tricky as they depend upon possible outcomes that are likely to occur for a particular event. Class 11 Probability will familiarize you with concepts and formulas that can help you with these predictions. The chapter covers a range of examples to help students understand the concept of Probability. This blog covers Probability study notes for class 11 along with some solved examples.

Class 11 Probability: Overview

The term probability is defined as a continuous measure of various uncertain events. It can be separated by using two prominent theories: the classical theory and the statistical approach of probability. As per class 11 maths syllabus, students can also explore probability in contrast with the axiomatic approach with suitable examples. The ultimate concept of class 11 chapter on probability is based on the formula:

Probability= Possible or average number of outcomes/ Total number of outcomes.

Class 11 Probability: Probability Conditions

These probability conditions are derived from the probability chapter and are really elaborative in nature so that you can easily prepare for the exams or while solving problems in general.

Condition: Probability Conditions: A Coin is Tossed Thrice

Solution:
A coin has 2 faces, one head (H) and another tail (T)
When the coin is tossed three times the actual number of its possible outcomes are 2×3=6 
So, when the coin is tossed thrice, it’s sample space is given as-
S =( HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)

Condition: A Coin is Tossed, and a Die is Thrown

Solution:
A coin has 2 faces, one head (H) and another tail (T)
A die has a total of 6 faces numbering from 1 to 6, and one number faces each face
So, when the coin is tossed, and a die is thrown the sample space is given as-
S =( H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6)

Condition: If the Dice is Thrown Twice

Solution:
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6
When a die is thrown twice its sample space is given by:
S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
The total no. of elements in this sample space => 6 × 6 = 36, while the sample space is given by:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3),(3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Condition: A Coin is Tossed Four Times

Solution:
Since either coin can turn up, Head (H) or Tail (T) are the possible outcomes. When 1 coin is tossed one time, its sample space is 2
Then, when the coin is tossed 3 times, the sample space is 24 = 16
Thus, the sample space,
S = {HHHH, THHH, HTHH, HHTH, HHHT, TTTT, HTTT, THTT, TTHT, TTTH, TTHH, HHTT, THTH, HTHT, THHT, HTTH}

Condition: Various Colours Dice
One dice of red colour, one of white colour, one of blue colour are placed in a bag. One die is selected at random and rolled, it’s colour, and the number on its uppermost face is noted. Discuss this sample space.

Solution:
Let us assume that 1, 2, 3, 4, 5, 6 are the possible numbers that come when the die is thrown
As per the class 11 chapter on probability, assume die of red colour be ‘R,’ die of white colour be ‘W,’ die of blue colour be ‘B.’ Hence, the total no. of sample space will be 6 x 3 = 18
The sample space of the event is
S={(R,1),(R,2),(R,3),(R,4),(R,5),(R,6),(W,1),(W,2),(W,3),(W,4),(W,5),(W,6) (B,1),(B,2),(B,3),(B,4),(B,5),(B,6)}

Note: ‘Die’ is singular in form whereas ‘Dice’ is plural in nature.

Class 11 Straight Lines

Class 11 Probability: Conditional Probability

The class 11 chapter on probability states that conditional probability and mutually exclusive events must be given due consideration from the examination point of view. The conditional probability represents the probability of an event taking place in a relationship to one or multiple events. For example, A and B are the two consecutive events associated with a similar sample space of any random experiment. The conditional probability for the event D will be under the condition where E has offered will be written as P (D | E)

Example: A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

Solution: Let us assume that 1, 2, 3, 4, 5, and 6 are the possible outcomes when the die is thrown.
As per the condition mentioned in the question, if a die is thrown repeatedly until six appears.
If six may come up for the first throw or six may come upon the second throw, this process will go continuously until the six comes.

The SS (sample space) when 6 appears on very first throw S1 = {6}
The SS (sample space) when 6 appears on second throw S2 = {(1,6), (2,6), (3,6), (4,6), (5,6)}
This event can go on for infinite times
So, the sample space is infinitely defined
S = {(6), (1,6), (2,6), (3,6), (4,6), (5,6), (1,1,6), (1,2,6)……}

Permutation and Combination Class 11

What Are Conditional Probability Independent Events?

For any two events, A and B associated with a sample space D. If the probability of the random occurrence of any one of the events is not affected by the other event’s occurrence, both of them are referred to as independent events. On the contrary, mutually exclusive events are the opposite representation of the same. A random variable refers to the function of a real-valued variable. The domain of this variable is the actual sample space of any random experiment.

Example: Let E and F be events with P (E) = 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5. Are E and F independent?

Solution: As per the class 11 maths NCERT solutions for this exercise-
P (E)= 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ≠ 1/5  
⇒ P (E ∩ F) ≠ P (E). P (F)

Class 11 Probability Sample Questions

1. If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

Ans: We have word ALGORITHM Number of letters = 9

2. A die is rolled in such a way that each odd number is twice as likely to occur as each even number.

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Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Ans: If is given that 2 x Probability of even number = Probability of odd number

3. Four candidates A, B, C, ZJhave applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that
(a) C will be selected? (b) A will not be selected?

Ans: It is given that A is twice as likely to be selected as B
P(A) = 2P(B)
B and C are given about the same chance of being selected
P(B) = P(C)
C is twice as likely to be selected as D
P(C) = 2 P(D)

4. Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

5. While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours

6. If 6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is

7. A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is

8. If the probabilities for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is

9. The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get a compartment is 0.96.

Answer: False
Let A = Student will pass examination
B = Student will getting compartment
P(A) = 0.73, P(B) = 0.13 and P(A or B) = 0.96
P(A or B) = P(A) + P(B) = 0.73 + 0.13 = 0.86
But P(A or B) = 0.96
Hence, the given statement is false.

10. The probability of intersection of two events A and B is always less than or equal to those favourable to the event

Answer: True
We know that A ∩ B ⊂ A
P(A ∩ B) ≤ P(A)
Hence, it is a true statement.

11. The probability of an occurrence of event A is .7 and that of the occurrence of event B is .3 and the probability of occurrence of both is .4.

Answer: False
A ∩B⊆ A, B
P(A ∩B) ≤ P(A), P(B)
But given that P(B) = 0.3 and P(A ∩B) = 0.4, which is not possible.

12. The sum of probabilities of two students getting distinction in their final examinations is 1.2.

Answer: True
Probability of each student getting distinction in their final examination is less than or equal to 1, the sum of the probabilities of two may be 1.2.
Hence, it is a true statement.

13. A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

Answer: In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2, 5, 6), … ,(5, 1, 6), (5, 2, 6), …}

14. A coin is tossed twice, what is the probability that at least one tail occurs?

Answer: When a coin is tossed twice, the sample space is given by
S = {HH, HT, TH, TT}
Let A be the event of the occurrence of at least one tail
Accordingly, A = {HT, TH, TT}

15. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

Answer: Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Accordingly, n(S) = 12
(i) Let A be the event in which the sum of numbers that turn up is 3
Accordingly, A = {(1, 2)}

(ii) Let B be the event in which the sum of numbers that turn up is 12
Accordingly, B = {(6, 6)}

Thus, we hope that after going through this blog on probability for class 11, you are now familiar with the topic. Our career experts at Leverage Edu are here to guide you making the right career choices in your dream field. Hurry up! Book an e-meeting with us!