Probability — Class 11 Maths Notes

Probability · Class 11 Maths · 12 topics.

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Topics covered in Probability

1.Introduction of Probability
Brief Explanation
Probability is a branch of mathematics that deals with calculating the likelihood of a given event's occurrence, which is expressed as a number between 1 and 0. Here, 1 indicates certainty and 0 indicates impossibility.
Detailed Explanation
What is Probability?
Probability helps us to quantify uncertainty. It's used to predict how likely an event is to happen. For example, when you flip a coin, there are two possible outcomes: heads or tails. Probability can tell us that the chance of getting heads is 50% (or 0.5).
Basic Terms in Probability:
1. Experiment: An action or process that leads to one or several outcomes. Example: Flipping a coin.
2. Outcome: A possible result of an experiment. Example: Heads or tails in a coin flip.
3. Event: A set of one or more outcomes. Example: Getting a heads in a coin flip.
4. Sample Space: The set of all possible outcomes. Example: {Heads, Tails} for a coin flip.
Calculating Probability
The probability of an event happening is calculated using the formula: Probability (P)=Number of favorable outcomesTotal number of possible outcomesProbability (P)=Total number of possible outcomesNumber of favorable outcomes
Example: If you roll a fair six-sided die, the probability of rolling a 4 is: 𝑃(4)=16P(4)=61 because there is only one favorable outcome (rolling a 4) out of six possible outcomes (1, 2, 3, 4, 5, 6).
Real-Life Example
Let's say you're playing a board game, and you need to roll a dice to decide your move. The game requires you to roll a 6 to win. The probability of rolling a 6 is: 𝑃(6)=16 P(6)=61
Careers and Industries Using Probability
1. Finance: Analysts use probability to assess risks and returns.
2. Insurance: Companies calculate the likelihood of events to set premiums.
3. Medicine: Probability helps in predicting the effectiveness of treatments.
4. Engineering: Reliability of systems and quality control use probability.
5. Sports: Strategies and outcomes are often based on probability.
Hands-On Activity
Try this simple activity at home to understand probability better:
1. Take a fair coin and flip it 10 times.
2. Record the outcomes (heads or tails) each time.
3. Calculate the probability of getting heads by using the formula: 𝑃(Heads)=Number of heads10 P(Heads)=10Number of heads
Repeat the experiment several times to see how the probability approaches 0.5.
Conclusion
Probability is a fascinating and useful field of mathematics that helps us make sense of uncertainty in everyday life and various professional fields.
2.Event
Brief Explanation
An event in probability is a specific outcome or a set of outcomes from a random experiment.
Detailed Explanation
What is an Event?
In probability, an event is defined as one or more outcomes of an experiment. An event can be as simple as a single outcome or as complex as a combination of multiple outcomes.
Types of Events:
1. Simple Event: An event that consists of exactly one outcome.
  - Example: Rolling a 3 on a die.
2. Compound Event: An event that consists of two or more outcomes.
  - Example: Rolling an even number on a die (outcomes: 2, 4, 6).
3. Certain Event: An event that is sure to happen.
  - Example: Getting a number between 1 and 6 when rolling a standard die.
4. Impossible Event: An event that cannot happen.
  - Example: Rolling a 7 on a standard six-sided die.
5. Mutually Exclusive Events: Events that cannot happen at the same time.
  - Example: Rolling a 2 and rolling a 5 on a single roll of a die.
6. Non-Mutually Exclusive Events: Events that can happen at the same time.
  - Example: Drawing a red card and drawing a king from a deck of cards (since there are red kings).
Example in Real Life
Consider a deck of cards. If you draw a card, the event of drawing a heart is a compound event since it includes all 13 hearts in the deck. Drawing the Ace of Spades is a simple event because it refers to just one specific card.
Calculating Probability of an Event
The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes in the sample space.
𝑃(Event)=Number of favorable outcomesTotal number of possible outcomesP(Event)=Total number of possible outcomesNumber of favorable outcomes
For example, if you want to find the probability of drawing a heart from a deck of 52 cards: 𝑃(Heart)=1352=14=0.25P(Heart)=5213=41=0.25
Careers and Industries Using Probability of Events
1. Gambling and Casinos: Probability is used to determine odds and payouts.
2. Weather Forecasting: Meteorologists predict the likelihood of various weather events.
3. Stock Market: Analysts use probability to forecast market movements and risks.
4. Quality Control: Engineers use probability to ensure product reliability.
5. Medicine: Doctors use probability to assess the likelihood of disease occurrences and treatment outcomes.
Hands-On Activity
To better understand events, try this simple activity:
1. Take a deck of cards and shuffle it.
2. Draw a card and note if it’s a heart, spade, club, or diamond.
3. Repeat this process 20 times, recording each draw.
4. Calculate the probability of drawing each suit by using the formula: 𝑃(Suit)=Number of times the suit was drawn20P(Suit)=20Number of times the suit was drawn
Compare your results with the theoretical probability of drawing each suit (which is 1441).
Conclusion
Understanding events in probability helps us make informed predictions and decisions in various aspects of life and different careers.
3.Occurrence of an Event
In probability and statistics, the "occurrence of an event" refers to a specific outcome or set of outcomes from an experiment or random process. An event is something that either happens or doesn't happen when an experiment is conducted. Let's understand this concept with simple examples and explanations.
Types of Events
There are several types of events in probability. Here are some of the most common ones:
1. Simple Event: An event with a single outcome.
  - Example: Rolling a die and getting a 3.
2. Compound Event: An event with two or more outcomes.
  - Example: Rolling a die and getting an even number (2, 4, or 6).
3. Certain Event: An event that will definitely happen.
  - Example: Rolling a die and getting a number less than 7.
4. Impossible Event: An event that cannot happen.
  - Example: Rolling a die and getting a number greater than 6.
5. Mutually Exclusive Events: Two events that cannot happen at the same time.
  - Example: Rolling a die and getting both an odd and an even number on the same roll.
6. Independent Events: The occurrence of one event does not affect the occurrence of another.
  - Example: Tossing a coin and rolling a die. The result of the coin toss does not affect the result of the die roll.
7. Dependent Events: The occurrence of one event affects the occurrence of another.
  - Example: Drawing two cards from a deck without replacement. The outcome of the first draw affects the second draw.
Detailed Explanation with an Example
Example: Tossing a Coin
Let's consider tossing a coin. When we toss a coin, there are two possible outcomes: Heads (H) or Tails (T). Here’s how different types of events apply:
1. Simple Event: Getting a Head (H) is a simple event.
2. Compound Event: Getting either a Head or a Tail (H or T) is a compound event.
3. Certain Event: Getting a Head or a Tail is a certain event because these are the only possible outcomes.
4. Impossible Event: Getting a number on the coin toss is an impossible event.
5. Mutually Exclusive Events: Getting a Head and getting a Tail are mutually exclusive events because they cannot happen simultaneously.
6. Independent Events: Tossing a coin twice. The result of the first toss does not affect the result of the second toss.
7. Dependent Events: This doesn’t apply directly to coin tosses but consider drawing cards from a deck. If you draw a card and do not put it back, the probability of drawing certain cards changes.
Real-Life Application
Understanding events and their types is crucial in fields such as:
- Finance: Analyzing the probability of stock price movements.
- Insurance: Calculating the likelihood of accidents or claims.
- Engineering: Assessing the probability of system failures.
- Medicine: Determining the chances of treatment success.
Activity
Toss a coin 20 times and record the outcome each time. Count how many times you get heads and how many times you get tails. This activity helps you understand simple, compound, and independent events.
4.Types of Events
1. Simple Event
A simple event is one that cannot be broken down into simpler components. It consists of a single outcome.
Example: Rolling a die and getting a 3.
2. Compound Event
A compound event consists of two or more simple events. It can happen in multiple ways.
Example: Rolling a die and getting an even number (2, 4, or 6).
3. Mutually Exclusive Events
Two events are mutually exclusive if they cannot happen at the same time.
Example: Rolling a die and getting a 3 or a 5. You cannot get both a 3 and a 5 on the same roll.
4. Exhaustive Events
A set of events is exhaustive if they cover all possible outcomes.
Example: Rolling a die and the events are getting a 1, 2, 3, 4, 5, or 6. These events cover all possible outcomes.
5. Independent Events
Two events are independent if the occurrence of one does not affect the occurrence of the other.
Example: Tossing a coin and rolling a die. The outcome of the coin toss does not affect the outcome of the die roll.
6. Dependent Events
Two events are dependent if the occurrence of one event affects the occurrence of the other.
Example: Drawing two cards from a deck without replacement. The outcome of the first draw affects the probability of the second draw.
Problem Solving Examples
Example 1: Mutually Exclusive Events
Problem: A box contains 3 red balls and 2 blue balls. One ball is drawn at random. What is the probability of drawing a red ball or a blue ball?
Solution:
- Probability of drawing a red ball 𝑃(𝑅)P(R) = 3553
- Probability of drawing a blue ball 𝑃(𝐵)P(B) = 2552
Since these are mutually exclusive events (drawing a red ball and a blue ball cannot happen at the same time), we can add their probabilities: 𝑃(𝑅 or 𝐵)=𝑃(𝑅)+𝑃(𝐵)=35+25=1P(R or B)=P(R)+P(B)=53+52=1
Example 2: Independent Events
Problem: A coin is tossed, and a die is rolled. What is the probability of getting a head on the coin and a 4 on the die?
Solution:
- Probability of getting a head 𝑃(𝐻)P(H) = 1221
- Probability of getting a 4 on the die 𝑃(4)P(4) = 1661
Since these are independent events, we multiply their probabilities: 𝑃(𝐻 and 4)=𝑃(𝐻)×𝑃(4)=12×16=112P(H and 4)=P(H)×P(4)=21×61=121
5.Exercise Questions
Question 1:
A die is rolled. Let E be the event "die shows 4" and F be the event "die shows an even number". Are E and F mutually exclusive?
Solution:
- Event E: The die shows 4. So, 𝐸={4}E={4}.
- Event F: The die shows an even number. So, 𝐹={2,4,6}F={2,4,6}.
Since 4 is an element of both events 𝐸E and 𝐹F, they are not mutually exclusive. They can occur at the same time.
Question 2:
A die is thrown. Describe the following events:
(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3
Solution:
(i) 𝐴={1,2,3,4,5,6}A={1,2,3,4,5,6}
(ii) 𝐵={}B={} (since there is no number greater than 7 on a die)
(iii) 𝐶={3,6}C={3,6}
(iv) 𝐷={1,2,3}D={1,2,3}
(v) 𝐸={6}E={6}
(vi) 𝐹={3,4,5,6}F={3,4,5,6}
Now, let's find the unions, intersections, and differences:
- 𝐴∪𝐵={1,2,3,4,5,6}∪{}={1,2,3,4,5,6}A∪B={1,2,3,4,5,6}∪{}={1,2,3,4,5,6}
- 𝐴∩𝐵={1,2,3,4,5,6}∩{}={}A∩B={1,2,3,4,5,6}∩{}={}
- 𝐵∪𝐶={}∪{3,6}={3,6}B∪C={}∪{3,6}={3,6}
- 𝐸∩𝐹={6}∩{3,4,5,6}={6}E∩F={6}∩{3,4,5,6}={6}
- 𝐷∩𝐸={1,2,3}∩{6}={}D∩E={1,2,3}∩{6}={}
- 𝐴−𝐶={1,2,3,4,5,6}−{3,6}={1,2,4,5}A−C={1,2,3,4,5,6}−{3,6}={1,2,4,5}
- 𝐷−𝐸={1,2,3}−{6}={1,2,3}D−E={1,2,3}−{6}={1,2,3}
- 𝐸∩𝐹′={6}∩{1,2,3,4,5}={}E∩F′={6}∩{1,2,3,4,5}={}
Question 3:
Three coins are tossed. Describe:
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.
Solution:
- When three coins are tossed, the possible outcomes are: {𝐻𝐻𝐻,𝐻𝐻𝑇,𝐻𝑇𝐻,𝐻𝑇𝑇,𝑇𝐻𝐻,𝑇𝐻𝑇,𝑇𝑇𝐻,𝑇𝑇𝑇}{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}.
(i) Two events which are mutually exclusive:
- Let 𝐴A be the event "getting 3 heads" ({HHH}).
- Let 𝐵B be the event "getting 3 tails" ({TTT}).
- 𝐴A and 𝐵B are mutually exclusive because both cannot happen at the same time.
(ii) Three events which are mutually exclusive and exhaustive:
- Let 𝐴A be the event "getting 0 heads" ({TTT}).
- Let 𝐵B be the event "getting 1 head" ({HTT, THT, TTH}).
- Let 𝐶C be the event "getting 2 heads" ({HHT, HTH, THH}).
- Let 𝐷D be the event "getting 3 heads" ({HHH}).
- 𝐴,𝐵,𝐶,𝐷A,B,C,D are mutually exclusive and exhaustive because they cover all possible outcomes.
(iii) Two events which are not mutually exclusive:
- Let 𝐴A be the event "getting at least 1 head" ({HHH, HHT, HTH, HTT, THH, THT, TTH}).
- Let 𝐵B be the event "getting at least 1 tail" ({HHT, HTH, HTT, THH, THT, TTH, TTT}).
- 𝐴A and 𝐵B are not mutually exclusive because they share common outcomes.
(iv) Two events which are mutually exclusive but not exhaustive:
- Let 𝐴A be the event "getting 3 heads" ({HHH}).
- Let 𝐵B be the event "getting 2 heads and 1 tail" ({HHT, HTH, THH}).
- 𝐴A and 𝐵B are mutually exclusive but not exhaustive because they do not cover all possible outcomes.
(v) Three events which are mutually exclusive but not exhaustive:
- Let 𝐴A be the event "getting 3 heads" ({HHH}).
- Let 𝐵B be the event "getting 3 tails" ({TTT}).
- Let 𝐶C be the event "getting 2 heads and 1 tail" ({HHT, HTH, THH}).
- 𝐴,𝐵,𝐶A,B,C are mutually exclusive but not exhaustive because they do not cover all possible outcomes.
Question 4:
Two dice are thrown. The events A, B and C are as follows:
- A: getting an even number on the first die.
- B: getting an odd number on the first die.
- C: getting the sum of the numbers on the dice ≤ 5.
Describe the events:
(i) A'
(ii) not B
(iii) A or B
(iv) A and B
(v) A but not C
(vi) B or C
(vii) B and C
(viii) A ∩ B' ∩ C'
Solution:
(i) 𝐴′A′ (not A): getting an odd number on the first die.
(ii) not Bnot B (B'): getting an even number on the first die.
(iii) 𝐴 or 𝐵A or B: getting either an even or odd number on the first die (this covers all possible outcomes).
(iv) 𝐴 and 𝐵A and B: There are no outcomes that can satisfy both A and B simultaneously, so 𝐴∩𝐵={}A∩B={}.
(v) 𝐴 but not 𝐶A but not C: getting an even number on the first die and the sum of the numbers on the dice > 5.
(vi) 𝐵 or 𝐶B or C: getting an odd number on the first die or the sum of the numbers on the dice ≤ 5.
(vii) 𝐵 and 𝐶B and C: getting an odd number on the first die and the sum of the numbers on the dice ≤ 5.
(viii) 𝐴∩𝐵′∩𝐶′A∩B′∩C′: getting an even number on the first die and the sum of the numbers on the dice > 5.
Question 5:
Refer to question 4 above, state true or false: (give reason for your answer)
(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) A = B'
(iv) A and C are mutually exclusive
(v) A and B' are mutually exclusive
(vi) 𝐴′,𝐵′,𝐶A′,B′,C are mutually exclusive and exhaustive
Solution:
(i) True. A and B are mutually exclusive because one cannot get both an even and an odd number on the same die roll.
(ii) True. A and B are mutually exclusive and exhaustive because together they cover all possible outcomes for the first die.
(iii) True. 𝐴A (getting an even number) is the complement of 𝐵B (getting an odd number).
(iv) False. A and C are not mutually exclusive because one can get an even number on the first die and still have the sum of the numbers on the dice ≤ 5.
(v) False. A and B' are not mutually exclusive because B' includes the same outcomes as A (getting an even number on the first die).
(vi) False. 𝐴′,𝐵′,𝐶A′,B′,C are not mutually exclusive and exhaustive. 𝐴′A′ and 𝐵′B′ cover all outcomes for the first die, but 𝐶C does not cover all possible sums for the two dice.
6.Axiomatic Approach to Probability
The axiomatic approach to probability, developed by the Russian mathematician Andrey Kolmogorov in 1933, forms the foundation of modern probability theory. This approach is based on a set of axioms (basic rules) that probability must satisfy.
Key Axioms of Probability:
1. Non-negativity: The probability of any event 𝐴A is a non-negative number.
  𝑃(𝐴)≥0P(A)≥0
2. Normalization: The probability of the sample space 𝑆S is 1.
  𝑃(𝑆)=1P(S)=1
3. Additivity: If 𝐴A and 𝐵B are two mutually exclusive events (they cannot happen simultaneously), then the probability of their union is the sum of their individual probabilities.
  𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)if 𝐴∩𝐵=∅P(A∪B)=P(A)+P(B)if A∩B=∅
These axioms are used to derive other properties and rules of probability.
Example to Understand Axioms
Imagine you have a fair six-sided die. The sample space 𝑆S for a roll of this die is:
𝑆={1,2,3,4,5,6}S={1,2,3,4,5,6}
1. Non-negativity: The probability of rolling any specific number, say 3, is non-negative.
  𝑃({3})=16≥0P({3})=61≥0
2. Normalization: The sum of probabilities of all possible outcomes is 1.
  𝑃({1})+𝑃({2})+𝑃({3})+𝑃({4})+𝑃({5})+𝑃({6})=16+16+16+16+16+16=1P({1})+P({2})+P({3})+P({4})+P({5})+P({6})=61+61+61+61+61+61=1
3. Additivity: If we want to find the probability of rolling a 2 or a 4, since these events are mutually exclusive (cannot happen at the same time), we add their probabilities.
  𝑃({2}∪{4})=𝑃({2})+𝑃({4})=16+16=13P({2}∪{4})=P({2})+P({4})=61+61=31
Examples Using the Axiomatic Approach to Probability
Example 1: Rolling Two Dice
Let's say we roll two six-sided dice and we want to find the probability of getting a sum of 7 or 11.
1. Identify the Sample Space: The sample space 𝑆S consists of all possible outcomes when two dice are rolled. There are 6×6=366×6=36 possible outcomes.
2. Events:
  - Let 𝐴A be the event that the sum is 7.
  - Let 𝐵B be the event that the sum is 11.
3. List Favorable Outcomes:
  - For 𝐴A (sum = 7): {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}. There are 6 outcomes.
  - For 𝐵B (sum = 11): {(5,6),(6,5)}{(5,6),(6,5)}. There are 2 outcomes.
4. Calculate Probabilities:
  𝑃(𝐴)=636=16P(A)=366=61𝑃(𝐵)=236=118P(B)=362=181
5. Check for Intersection:
  - 𝐴∩𝐵A∩B: Since it’s impossible for a sum to be both 7 and 11 at the same time, 𝑃(𝐴∩𝐵)=0P(A∩B)=0.
6. Apply the Addition Rule:
  𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)P(A∪B)=P(A)+P(B)−P(A∩B)𝑃(𝐴∪𝐵)=16+118−0=318+118=418=29P(A∪B)=61+181−0=183+181=184=92
So, the probability of getting a sum of 7 or 11 when rolling two dice is 2992.
Example 2: Drawing Cards from a Deck
Suppose you draw one card from a standard deck of 52 cards. What is the probability of drawing either a King or a Heart?
1. Identify the Sample Space: The sample space 𝑆S consists of all 52 cards in the deck.
2. Events:
  - Let 𝐴A be the event that the card is a King.
  - Let 𝐵B be the event that the card is a Heart.
3. List Favorable Outcomes:
  - For 𝐴A (King): There are 4 Kings in a deck.
  - For 𝐵B (Heart): There are 13 Hearts in a deck.
4. Calculate Probabilities:
  𝑃(𝐴)=452=113P(A)=524=131𝑃(𝐵)=1352=14P(B)=5213=41
5. Check for Intersection:
  - 𝐴∩𝐵A∩B: There is 1 card that is both a King and a Heart (King of Hearts).
  𝑃(𝐴∩𝐵)=152P(A∩B)=521
6. Apply the Addition Rule:
  𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)P(A∪B)=P(A)+P(B)−P(A∩B)𝑃(𝐴∪𝐵)=113+14−152P(A∪B)=131+41−521
  To add these fractions, find a common denominator (52 in this case):
  𝑃(𝐴∪𝐵)=452+1352−152=1652=413P(A∪B)=524+5213−521=5216=134
So, the probability of drawing either a King or a Heart from a standard deck of 52 cards is 413134
.Real-Life Application
Probability theory is widely used in various fields such as finance, insurance, medicine, and engineering. For example, in finance, the axiomatic approach helps in modeling the likelihood of various market events, enabling risk assessment and decision-making.
Careers Using Probability:
- Statisticians: Analyze data and apply probability to make predictions.
- Actuaries: Use probability to assess risks in insurance and finance.
- Data Scientists: Utilize probability for data analysis and machine learning.
- Financial Analysts: Apply probability to forecast market trends and evaluate investments.
Step-by-Step Derivation of an Important Formula: Addition Rule
The addition rule for any two events 𝐴A and 𝐵B states:
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)P(A∪B)=P(A)+P(B)−P(A∩B)
Step-by-Step Explanation:
1. Start with the events: Let 𝐴A and 𝐵B be any two events in a sample space 𝑆S.
2. Mutually Exclusive Events: If 𝐴A and 𝐵B are mutually exclusive, 𝐴∩𝐵=∅A∩B=∅, then:
  𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)P(A∪B)=P(A)+P(B)
3. Non-Mutually Exclusive Events: For events that are not mutually exclusive, some outcomes are counted twice (once in 𝑃(𝐴)P(A) and once in 𝑃(𝐵)P(B)). To correct this, we subtract the intersection:
  𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)P(A∪B)=P(A)+P(B)−P(A∩B)
7.Probability of an Event
Probability is a measure of how likely an event is to occur. It is expressed as a number between 0 and 1, where 0 means the event will not occur and 1 means the event will certainly occur.
Basic Definition of Probability
For a given event 𝐴A, the probability 𝑃(𝐴)P(A) is defined as:
𝑃(𝐴)=Number of favorable outcomesTotal number of possible outcomesP(A)=Total number of possible outcomesNumber of favorable outcomes
Example 1: Flipping a Coin
Consider the simple case of flipping a fair coin. The sample space 𝑆S is:
𝑆={Heads, Tails}S={Heads, Tails}
1. Event 𝐴A: Getting a Heads.
2. Favorable Outcomes: 1 (Heads)
3. Total Possible Outcomes: 2 (Heads, Tails)
The probability 𝑃(𝐴)P(A) is:
𝑃(𝐴)=12P(A)=21
So, the probability of getting Heads when flipping a fair coin is 1221 or 0.5.
Example 2: Drawing a Card from a Deck
Let's say you draw one card from a standard deck of 52 cards. What is the probability of drawing an Ace?
1. Event 𝐴A: Drawing an Ace.
2. Favorable Outcomes: 4 (one Ace each of hearts, diamonds, clubs, and spades)
3. Total Possible Outcomes: 52 (all the cards in the deck)
The probability 𝑃(𝐴)P(A) is:
𝑃(𝐴)=452=113P(A)=524=131
So, the probability of drawing an Ace from a standard deck of cards is 113131 or approximately 0.077.
Real-Life Applications
Probability is used extensively in various fields:
- Medicine: To determine the likelihood of a patient developing a disease.
- Weather Forecasting: To predict the chance of rain or other weather conditions.
- Insurance: To calculate the risk and set premiums for insurance policies.
- Gaming: To design fair games and understand the odds of winning.
- Finance: To assess risks and returns on investments.
Careers Using Probability:
- Data Analyst: Uses probability to interpret data and make predictions.
- Actuary: Applies probability to assess risks and design insurance policies.
- Economist: Uses probability to model economic behaviors and forecast trends.
- Engineer: Applies probability in quality control and reliability testing.
- Meteorologist: Uses probability to predict weather patterns.
8.Probabilities of Equally Likely Outcomes
What is Probability?
Probability is a measure of the likelihood of an event occurring. It ranges from 0 to 1, where 0 means the event cannot happen and 1 means the event is certain to happen.
Equally Likely Outcomes
When we say that outcomes are "equally likely," it means that each outcome has the same chance of occurring. To calculate the probability of an event with equally likely outcomes, we use the formula:
Probability=Number of favorable outcomesTotal number of possible outcomesProbability=Total number of possible outcomesNumber of favorable outcomes
Example 1: Spinning a Spinner
Imagine you have a spinner divided into 8 equal sections, numbered 1 through 8. Each section is equally likely to land under the pointer when the spinner stops.
- Total number of possible outcomes: 8 (since there are 8 sections)
- Number of favorable outcomes for landing on a number greater than 6: 2 (sections 7 and 8)
Using the probability formula:
Probability of landing on a number greater than 6=28=14Probability of landing on a number greater than 6=82=41
Example 2: Selecting a T-shirt from a Drawer
Suppose you have a drawer with 10 T-shirts of different colors: 4 red, 3 blue, 2 green, and 1 yellow. Each T-shirt is equally likely to be selected.
- Total number of possible outcomes: 10 (total T-shirts)
- Number of favorable outcomes for selecting a red T-shirt: 4
Using the probability formula:
Probability of selecting a red T-shirt=410=25Probability of selecting a red T-shirt=104=52
Example 3: Drawing a Token from a Jar
Imagine you have a jar containing 12 tokens numbered 1 to 12. Each token is equally likely to be drawn.
- Total number of possible outcomes: 12 (since there are 12 tokens)
- Number of favorable outcomes for drawing an even-numbered token: 6 (tokens 2, 4, 6, 8, 10, 12)
Using the probability formula:
Probability of drawing an even-numbered token=612=12Probability of drawing an even-numbered token=126=21
Summary
In each of these examples, we calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. This method can be applied to any situation where the outcomes are equally likely.
9.Probability of the Event 'A or B'
What is the Probability of the event 'A or B'?
The probability of the event 'A or B' occurring is the probability that either event A occurs, event B occurs, or both events A and B occur. This is calculated using the formula for the union of two events.
Formula for the Probability of 'A or B'
For any two events A and B, the probability of 'A or B' (denoted as 𝑃(𝐴∪𝐵)P(A∪B)) is given by:
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)P(A∪B)=P(A)+P(B)−P(A∩B)
Here,
- 𝑃(𝐴)P(A) is the probability of event A.
- 𝑃(𝐵)P(B) is the probability of event B.
- 𝑃(𝐴∩𝐵)P(A∩B) is the probability that both events A and B occur (the intersection of A and B).
Why Subtract 𝑃(𝐴∩𝐵)P(A∩B)?
We subtract 𝑃(𝐴∩𝐵)P(A∩B) because when we add 𝑃(𝐴)P(A) and 𝑃(𝐵)P(B), the overlap (where both A and B occur) is counted twice. To correct this, we subtract the intersection 𝑃(𝐴∩𝐵)P(A∩B).
Example 1: Rolling a Die
Consider rolling a fair six-sided die. Let event A be rolling an even number, and event B be rolling a number greater than 4.
- Event A (rolling an even number): Possible outcomes are {2, 4, 6}. 𝑃(𝐴)=36=12P(A)=63=21
- Event B (rolling a number greater than 4): Possible outcomes are {5, 6}. 𝑃(𝐵)=26=13P(B)=62=31
- Intersection of A and B (rolling a 6): Possible outcome is {6}. 𝑃(𝐴∩𝐵)=16P(A∩B)=61
Using the formula:
𝑃(𝐴∪𝐵)=12+13−16=36+26−16=46=23P(A∪B)=21+31−61=63+62−61=64=32
Example 2: Drawing a Card from a Deck
Consider drawing a card from a standard deck of 52 cards. Let event A be drawing a heart, and event B be drawing a face card (Jack, Queen, King).
- Event A (drawing a heart): There are 13 hearts. 𝑃(𝐴)=1352=14P(A)=5213=41
- Event B (drawing a face card): There are 12 face cards. 𝑃(𝐵)=1252=313P(B)=5212=133
- Intersection of A and B (drawing a face card that is a heart): There are 3 face cards that are hearts. 𝑃(𝐴∩𝐵)=352P(A∩B)=523
Using the formula:
𝑃(𝐴∪𝐵)=14+313−352P(A∪B)=41+133−523
To calculate, we first convert the fractions to a common denominator (52):
𝑃(𝐴∪𝐵)=1352+1252−352=2252=1126P(A∪B)=5213+5212−523=5222=2611
Example 3: Flipping Two Coins
Consider flipping two fair coins. Let event A be getting at least one head, and event B be getting two tails.
- Event A (at least one head): Possible outcomes are {HT, TH, HH}. 𝑃(𝐴)=34P(A)=43
- Event B (two tails): Possible outcome is {TT}. 𝑃(𝐵)=14P(B)=41
- Intersection of A and B (getting both heads and two tails): This is impossible. 𝑃(𝐴∩𝐵)=0P(A∩B)=0
Using the formula:
𝑃(𝐴∪𝐵)=34+14−0=1P(A∪B)=43+41−0=1
This makes sense because the events cover all possible outcomes.
Summary
The formula 𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)P(A∪B)=P(A)+P(B)−P(A∩B) allows us to calculate the probability of either event A or event B occurring by adding the individual probabilities and subtracting the overlap. This approach ensures that we do not double-count the intersection of the two events.
10.Probability of the Event 'Not A'
What is the Probability of 'Not A'?
The probability of the event 'Not A' (denoted as 𝐴‾A or 𝐴′A′) is the probability that event A does not occur. It is calculated as:
𝑃(𝐴‾)=1−𝑃(𝐴)P(A)=1−P(A)
This formula comes from the fact that the total probability of all possible outcomes in a probability space is 1. So, if we know the probability of event A occurring, we can subtract it from 1 to find the probability of event A not occurring.
Example 1: Drawing a Red Ball from a Bag
Imagine you have a bag with 7 red balls and 3 blue balls. Each ball is equally likely to be drawn.
- Total number of balls: 10 (7 red + 3 blue)
- Number of red balls (event A): 7
First, we calculate the probability of drawing a red ball (event A):
𝑃(𝐴)=710P(A)=107
Now, we calculate the probability of not drawing a red ball (event 𝐴‾A):
𝑃(𝐴‾)=1−𝑃(𝐴)=1−710=310P(A)=1−P(A)=1−107=103
Example 2: Rolling a Die
Consider rolling a fair six-sided die. Let event A be rolling a number greater than 4 (i.e., rolling a 5 or 6).
- Total number of possible outcomes: 6 (since there are 6 sides)
- Number of favorable outcomes for event A: 2 (rolling a 5 or 6)
First, we calculate the probability of rolling a number greater than 4 (event A):
𝑃(𝐴)=26=13P(A)=62=31
Now, we calculate the probability of not rolling a number greater than 4 (event 𝐴‾A):
𝑃(𝐴‾)=1−𝑃(𝐴)=1−13=23P(A)=1−P(A)=1−31=32
Example 3: Choosing a Student at Random
Assume a class has 20 students: 8 boys and 12 girls. Let event A be choosing a girl.
- Total number of students: 20 (8 boys + 12 girls)
- Number of girls (event A): 12
First, we calculate the probability of choosing a girl (event A):
𝑃(𝐴)=1220=35P(A)=2012=53
Now, we calculate the probability of not choosing a girl (event 𝐴‾A):
𝑃(𝐴‾)=1−𝑃(𝐴)=1−35=25P(A)=1−P(A)=1−53=52
Summary
The probability of the event 'Not A' is calculated by subtracting the probability of event A from 1. This gives us the likelihood of event A not occurring.
11.Exercise Questions
Question 1:
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
Solution:
- Coin outcomes: {1, 6}
- Die outcomes: {1, 2, 3, 4, 5, 6}
Let's find the required probabilities for each sum:
(i) Sum = 3:
- Possible outcomes: (1 on coin, 2 on die)
Total possible outcomes for the coin and die: 2 (coin outcomes)×6 (die outcomes)=122 (coin outcomes)×6 (die outcomes)=12
Number of favorable outcomes:
- (1 on coin, 2 on die)
𝑃(sum = 3)=Number of favorable outcomesTotal number of outcomes=112P(sum = 3)=Total number of outcomesNumber of favorable outcomes=121
(ii) Sum = 12:
- Possible outcomes: (6 on coin, 6 on die)
Number of favorable outcomes:
- (6 on coin, 6 on die)
𝑃(sum = 12)=Number of favorable outcomesTotal number of outcomes=112P(sum = 12)=Total number of outcomesNumber of favorable outcomes=121
Question 2:
There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Solution:
- Total number of council members: 4 men + 6 women = 10
- Number of women: 6
Probability of selecting a woman: 𝑃(woman)=Number of womenTotal number of council members=610=35P(woman)=Total number of council membersNumber of women=106=53
Question 3:
A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
Solution:
First, let's determine the different amounts of money based on the possible outcomes of the coin tosses.
- HHHH: 4 heads, 0 tails Earnings=4×1=4Earnings=4×1=4
- HHHT, HHTH, HTHH, THHH: 3 heads, 1 tail Earnings=3×1−1.5=3−1.5=1.5Earnings=3×1−1.5=3−1.5=1.5
- HHTT, HTHT, HTTH, TTHH, THTH, THHT: 2 heads, 2 tails Earnings=2×1−2×1.5=2−3=−1Earnings=2×1−2×1.5=2−3=−1
- HTTT, THTT, TTHT, TTTH: 1 head, 3 tails Earnings=1×1−3×1.5=1−4.5=−3.5Earnings=1×1−3×1.5=1−4.5=−3.5
- TTTT: 0 heads, 4 tails Earnings=0×1−4×1.5=0−6=−6Earnings=0×1−4×1.5=0−6=−6
Different amounts of money and their probabilities:
1. Rs 4
  - Only one way: HHHH
  - Probability: 𝑃(Rs 4)=116P(Rs 4)=161
2. Rs 1.5
  - Four ways: HHHT, HHTH, HTHH, THHH
  - Probability: 𝑃(Rs 1.5)=416=14P(Rs 1.5)=164=41
3. Rs -1
  - Six ways: HHTT, HTHT, HTTH, TTHH, THTH, THHT
  - Probability: 𝑃(Rs -1)=616=38P(Rs -1)=166=83
4. Rs -3.5
  - Four ways: HTTT, THTT, TTHT, TTTH
  - Probability: 𝑃(Rs -3.5)=416=14P(Rs -3.5)=164=41
5. Rs -6
  - Only one way: TTTT
  - Probability: 𝑃(Rs -6)=116P(Rs -6)=161
12.Exercise Questions
Question 01:
In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Solution:
Let:
- 𝑃(𝑀)P(M) = Probability of studying Mathematics = 0.40
- 𝑃(𝐵)P(B) = Probability of studying Biology = 0.30
- 𝑃(𝑀∩𝐵)P(M∩B) = Probability of studying both Mathematics and Biology = 0.10
We need to find the probability of a student studying Mathematics or Biology, 𝑃(𝑀∪𝐵)P(M∪B).
Using the formula for the union of two events: 𝑃(𝑀∪𝐵)=𝑃(𝑀)+𝑃(𝐵)−𝑃(𝑀∩𝐵)P(M∪B)=P(M)+P(B)−P(M∩B)
Substituting the given values: 𝑃(𝑀∪𝐵)=0.40+0.30−0.10=0.60P(M∪B)=0.40+0.30−0.10=0.60
So, the probability that a randomly selected student will be studying Mathematics or Biology is 0.60.
Question 02:
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Solution:
Let:
- 𝑃(𝐴)P(A) = Probability of passing the first examination = 0.80
- 𝑃(𝐵)P(B) = Probability of passing the second examination = 0.70
- 𝑃(𝐴∪𝐵)P(A∪B) = Probability of passing at least one examination = 0.95
We need to find the probability of passing both examinations, 𝑃(𝐴∩𝐵)P(A∩B).
Using the formula for the union of two events: 𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)P(A∪B)=P(A)+P(B)−P(A∩B)
Rearranging to find 𝑃(𝐴∩𝐵)P(A∩B): 𝑃(𝐴∩𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∪𝐵)P(A∩B)=P(A)+P(B)−P(A∪B)
Substituting the given values: 𝑃(𝐴∩𝐵)=0.80+0.70−0.95=0.55P(A∩B)=0.80+0.70−0.95=0.55
So, the probability of passing both examinations is 0.55.
Question 03:
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?
Solution:
Let:
- 𝑃(𝐸)P(E) = Probability of passing English examination = 0.75
- 𝑃(𝐻)P(H) = Probability of passing Hindi examination
- 𝑃(𝐸∩𝐻)P(E∩H) = Probability of passing both English and Hindi = 0.50
- 𝑃(𝐸‾∩𝐻‾)P(E∩H) = Probability of passing neither = 0.10
We need to find 𝑃(𝐻)P(H).
Using the complement rule: 𝑃(𝐸‾∩𝐻‾)=1−𝑃(𝐸∪𝐻)P(E∩H)=1−P(E∪H)
So, 𝑃(𝐸∪𝐻)=1−𝑃(𝐸‾∩𝐻‾)=1−0.10=0.90P(E∪H)=1−P(E∩H)=1−0.10=0.90
Using the formula for the union of two events: 𝑃(𝐸∪𝐻)=𝑃(𝐸)+𝑃(𝐻)−𝑃(𝐸∩𝐻)P(E∪H)=P(E)+P(H)−P(E∩H)
Rearranging to find 𝑃(𝐻)P(H): 𝑃(𝐻)=𝑃(𝐸∪𝐻)+𝑃(𝐸∩𝐻)−𝑃(𝐸)P(H)=P(E∪H)+P(E∩H)−P(E)
Substituting the given values: 𝑃(𝐻)=0.90+0.50−0.75=0.65P(H)=0.90+0.50−0.75=0.65
So, the probability of passing the Hindi examination is 0.65.
Question: 04
In a class of 60 students, 30 opted for NCC, 32 opted for NSS, and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that:
(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.
Given Data:
- Total number of students, 𝑁N = 60
- Students who opted for NCC, 𝑛(𝑁𝐶𝐶)n(NCC) = 30
- Students who opted for NSS, 𝑛(𝑁𝑆𝑆)n(NSS) = 32
- Students who opted for both NCC and NSS, 𝑛(𝑁𝐶𝐶∩𝑁𝑆𝑆)n(NCC∩NSS) = 24
Solution:
(i) Probability that the student opted for NCC or NSS:
We need to find 𝑃(𝑁𝐶𝐶∪𝑁𝑆𝑆)P(NCC∪NSS).
Using the formula for the union of two sets: 𝑃(𝑁𝐶𝐶∪𝑁𝑆𝑆)=𝑃(𝑁𝐶𝐶)+𝑃(𝑁𝑆𝑆)−𝑃(𝑁𝐶𝐶∩𝑁𝑆𝑆)P(NCC∪NSS)=P(NCC)+P(NSS)−P(NCC∩NSS)
First, we find the probabilities: 𝑃(𝑁𝐶𝐶)=𝑛(𝑁𝐶𝐶)𝑁=3060=12P(NCC)=Nn(NCC)=6030=21 𝑃(𝑁𝑆𝑆)=𝑛(𝑁𝑆𝑆)𝑁=3260=815P(NSS)=Nn(NSS)=6032=158 𝑃(𝑁𝐶𝐶∩𝑁𝑆𝑆)=𝑛(𝑁𝐶𝐶∩𝑁𝑆𝑆)𝑁=2460=25P(NCC∩NSS)=Nn(NCC∩NSS)=6024=52
Now, substitute these into the formula: 𝑃(𝑁𝐶𝐶∪𝑁𝑆𝑆)=12+815−25P(NCC∪NSS)=21+158−52
Finding a common denominator (which is 30 in this case): 𝑃(𝑁𝐶𝐶∪𝑁𝑆𝑆)=1530+1630−1230=1930P(NCC∪NSS)=3015+3016−3012=3019
So, the probability that the student opted for NCC or NSS is 19303019.
(ii) Probability that the student has opted neither NCC nor NSS:
This is the complement of the probability that the student opted for NCC or NSS.
𝑃(neither NCC nor NSS)=1−𝑃(𝑁𝐶𝐶∪𝑁𝑆𝑆)P(neither NCC nor NSS)=1−P(NCC∪NSS)
From part (i), we know: 𝑃(𝑁𝐶𝐶∪𝑁𝑆𝑆)=1930P(NCC∪NSS)=3019
So, 𝑃(neither NCC nor NSS)=1−1930=3030−1930=1130P(neither NCC nor NSS)=1−3019=3030−3019=3011
So, the probability that the student has opted neither NCC nor NSS is 11303011.
(iii) Probability that the student has opted NSS but not NCC:
We need to find the number of students who opted for NSS but not NCC, which is 𝑛(𝑁𝑆𝑆)−𝑛(𝑁𝐶𝐶∩𝑁𝑆𝑆)n(NSS)−n(NCC∩NSS).
𝑛(𝑁𝑆𝑆∖𝑁𝐶𝐶)=𝑛(𝑁𝑆𝑆)−𝑛(𝑁𝐶𝐶∩𝑁𝑆𝑆)=32−24=8n(NSS∖NCC)=n(NSS)−n(NCC∩NSS)=32−24=8
So, the probability is: 𝑃(𝑁𝑆𝑆∖𝑁𝐶𝐶)=𝑛(𝑁𝑆𝑆∖𝑁𝐶𝐶)𝑁=860=215P(NSS∖NCC)=Nn(NSS∖NCC)=608=152
So, the probability that the student has opted NSS but not NCC is 215152.

Probability — Class 11 Maths Notes

Topics covered in Probability

1.Introduction of Probability

Brief Explanation

Detailed Explanation

What is Probability?

Basic Terms in Probability:

Calculating Probability

Real-Life Example

Careers and Industries Using Probability

Hands-On Activity

Conclusion

2.Event

Brief Explanation

Detailed Explanation

What is an Event?

Types of Events:

Example in Real Life

Calculating Probability of an Event

Careers and Industries Using Probability of Events

Hands-On Activity

Conclusion

3.Occurrence of an Event

Types of Events

Detailed Explanation with an Example

Real-Life Application

Activity

4.Types of Events

1. Simple Event

2. Compound Event

3. Mutually Exclusive Events

4. Exhaustive Events

5. Independent Events

6. Dependent Events

Problem Solving Examples

Example 1: Mutually Exclusive Events

Example 2: Independent Events

5.Exercise Questions

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

6.Axiomatic Approach to Probability

The axiomatic approach to probability, developed by the Russian mathematician Andrey Kolmogorov in 1933, forms the foundation of modern probability theory. This approach is based on a set of axioms (basic rules) that probability must satisfy.

Key Axioms of Probability:

Example to Understand Axioms

Examples Using the Axiomatic Approach to Probability

Example 1: Rolling Two Dice

Example 2: Drawing Cards from a Deck

.Real-Life Application

Careers Using Probability:

Step-by-Step Derivation of an Important Formula: Addition Rule

7.Probability of an Event

Probability is a measure of how likely an event is to occur. It is expressed as a number between 0 and 1, where 0 means the event will not occur and 1 means the event will certainly occur.

Basic Definition of Probability

Example 1: Flipping a Coin

Example 2: Drawing a Card from a Deck

Real-Life Applications

Careers Using Probability:

8.Probabilities of Equally Likely Outcomes

What is Probability?

Equally Likely Outcomes

Example 1: Spinning a Spinner

Example 2: Selecting a T-shirt from a Drawer

Example 3: Drawing a Token from a Jar

Summary

9.Probability of the Event 'A or B'

What is the Probability of the event 'A or B'?

Formula for the Probability of 'A or B'

Why Subtract 𝑃(𝐴∩𝐵)P(A∩B)?

Example 1: Rolling a Die

Example 2: Drawing a Card from a Deck

Example 3: Flipping Two Coins

Summary

10.Probability of the Event 'Not A'

What is the Probability of 'Not A'?

Example 1: Drawing a Red Ball from a Bag

Example 2: Rolling a Die

Example 3: Choosing a Student at Random