Probability .

The mathematics of uncertainty. From coin tosses to Bayes' prediction models.

Random Experiments & Events

Key Terms

Random Experiment:

An experiment whose outcome cannot be predicted with certainty (e.g., Tossing a coin).

Sample Space (S):

The set of all possible outcomes of a random experiment. Ex: Coin toss

S = \{H, T\}

Event (E):

Any subset of the sample space S. Ex: Getting a Head

E = \{H\}

Mutually Exclusive:

Events that cannot happen simultaneously.

A \cap B = \phi

Exhaustive Events:

Set of events whose union is the sample space.

E_1 \cup E_2 \cup ... = S

Types of Events

Impossible Event

P(\phi) = 0

Probability is 0. Ex: Rolling a 7 on a die.

Sure Event

P(S) = 1

Probability is 1. Ex: Rolling number < 7 on a die.

Simple Event

E = \{w_i\}

Has only one sample point.

Compound Event

Has more than one sample point.

Algebra of Probability

Understanding relationships between events using Set Theory (Union, Intersection, Complement).

Visualizing Probability

Formula & Concept

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Addition Theorem

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Example:

P(A)=0.6, P(B)=0.4, P(A \cap B)=0.2

, find

P(A \cup B)

P(A \cup B) = 0.6 + 0.4 - 0.2 = 0.8

Axiomatic Definition

Probability $P(E)$ is a real-valued function satisfying:

$0 \le P(E) \le 1$
$P(S) = 1$
If $A, B$ are mutually exclusive, $P(A \cup B) = P(A) + P(B)$

Conditional Probability & Bayes

Conditional Probability

Probability of event A happening given that event B has already happened.

Formula

P(A|B) = \frac{P(A \cap B)}{P(B)}, \text{ provided } P(B) \neq 0

Multiplication Thm

P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)

Properties

$P(S|F) = P(F|F) = 1$
$P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$
$P(E'|F) = 1 - P(E|F)$

Total Probability & Bayes Theorem

Law of Total Probability

Total probability of A occurring is sum of all paths leading to A:

P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)

Bayes Theorem (Reverse)

Given A happened, chance it came from path E1?

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

Example: Total Probability

Bag I: 3 Red, 4 Black. Bag II: 5 Red, 6 Black. One bag is selected at random and a ball is drawn. Find prob it is Red.

P(R) = P(E_1)P(R|E_1) + P(E_2)P(R|E_2)

= \frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}

= \frac{3}{14} + \frac{5}{22} = \frac{33+35}{154} = \frac{68}{154} = \frac{34}{77}

Example: Bayes Theorem

Bag I: 3 Red, 4 Black. Bag II: 5 Red, 6 Black. One ball drawn from selected bag is Red. Find prob it is from Bag II.

Identify the reverse probability question: "Given Outcome (Red), find Source (Bag II)". This is a classic Bayes signature.

1. Events:

E_1

(Bag I),

E_2

(Bag II),

A

(Red Ball) 2. Priors:

P(E_1)=P(E_2)=1/2

3. Conditionals:

P(A|E_1)=3/7

(Red from Bag I),

P(A|E_2)=5/11

(Red from Bag II) 4. Bayes Formula:

P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

5. Calc:

\frac{1/2 \cdot 5/11}{1/2 \cdot 3/7 + 1/2 \cdot 5/11} = \frac{5/22}{34/77} = \frac{35}{68}

✏️ Practice Zone

Problems

A coin is tossed three times. What is the probability of getting at least two heads?

✓ Show Solution

1. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8

2. E (at least 2 H) = {HHH, HHT, HTH, THH}, n(E) = 4

P(E) = 4/8 = 1/2

EASY

If A and B are two events such that

P(A) = 1/4, P(B) = 1/2

and

P(A \cap B) = 1/8

, find

P( ext{not A and not B})

✓ Show Solution

P(A' \cap B') = P((A \cup B)')

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1/4 + 1/2 - 1/8 = (2+4-1)/8 = 5/8

P(A' \cap B') = 1 - 5/8 = 3/8

MEDIUM

A die is thrown twice and the sum of numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

✓ Show Solution

1. E (sum 6) = {(1,5), (2,4), (3,3), (4,2), (5,1)}. n(E)=5.

2. F (4 at least once) = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}.

E \cap F = \{(2,4), (4,2)\}

n(E \cap F) = 2

P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{2/36}{5/36} = 2/5

MEDIUM

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

✓ Show Solution

P(B_1) = 26/52 = 1/2

P(B_2|B_1) = 25/51

(since one black card is gone)

P(B_1 \cap B_2) = P(B_1)P(B_2|B_1) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}

EASY

In a class, 40% students study Math, 25% study Bio and 15% study both. One student is selected at random. (i) If he studies Math, what is prob he studies Bio? (ii) If he studies Bio, prob he studies Math?

✓ Show Solution

P(M)=0.4, P(B)=0.25, P(M \cap B)=0.15

2. (i)

P(B|M) = \frac{P(B \cap M)}{P(M)} = \frac{0.15}{0.40} = \frac{15}{40} = 3/8

3. (ii)

P(M|B) = \frac{P(M \cap B)}{P(B)} = \frac{0.15}{0.25} = \frac{15}{25} = 3/5

MEDIUM

A die is tossed thrice. Find the probability of getting an odd number at least once.

✓ Show Solution

1. P(Odd) in one toss = 3/6 = 1/2. P(Even) = 1/2.

2. P(At least one Odd) = 1 - P(None Odd) = 1 - P(All Even)

3. Since tosses independent: P(All Even) =

P(E)P(E)P(E) = (1/2)^3 = 1/8

4. Ans:

1 - 1/8 = 7/8

MEDIUM

CBSE Exam Corner

Master these questions to ace your board exams.

CBSE SampleID: prob_exam_1

Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

Solution Strategy1. Let K, K, A be events. 2. P(K1) = 4/52 3. P(K2|K1) = 3/51 4. P(A|K1K2) = 4/50 5. Total Prob =

\frac{4}{52} \times \frac{3}{51} \times \frac{4}{50} = \frac{1}{13} \times \frac{1}{17} \times \frac{2}{25} = \frac{2}{5525}

CBSE 2023ID: prob_exam_2

In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective type. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

Solution Strategy1. Let

E_1, E_2, E_3

be events of Machine A, B, C.

P(E_1)=0.25, P(E_2)=0.35, P(E_3)=0.40

. 2. Let D be defective.

P(D|E_1)=0.05, P(D|E_2)=0.04, P(D|E_3)=0.02

. 3. Bayes:

P(E_2|D) = \frac{P(E_2)P(D|E_2)}{\sum P(E_i)P(D|E_i)}

4. Numerator:

0.35 \times 0.04 = 0.0140

5. Denom:

0.25(0.05) + 0.35(0.04) + 0.40(0.02) = 0.0125 + 0.0140 + 0.0080 = 0.0345

6. Ans:

\frac{0.0140}{0.0345} = \frac{140}{345} = 28/69

CBSE ModelID: prob_exam_3

A and B toss a coin alternately till one of them tosses a head and wins the game. If A starts the game, find their respective probabilities of winning.

Solution Strategy1. P(H) = 1/2, P(T) = 1/2. 2. A wins in 1st (H), 3rd (TTH), 5th (TTTTH)... toss. 3. P(A wins) =

1/2 + (1/2)^3 + (1/2)^5 + ...

4. This is a GP with

a=1/2, r=1/4

. 5. Sum =

\frac{a}{1-r} = \frac{1/2}{1-1/4} = \frac{1/2}{3/4} = 2/3

. 6. P(B wins) =

1 - P(A) = 1 - 2/3 = 1/3

ImportantID: prob_exam_4

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 solves the problem.

Solution Strategy1.

P(A)=1/2, P(B)=1/3

P(A')=1/2, P(B')=2/3

. 2. (i) Prob problem solved =

P(A \cup B) = 1 - P(A' \cap B') = 1 - P(A')P(B') = 1 - (1/2)(2/3) = 1 - 1/3 = 2/3

. 3. (ii) Exactly one =

P(A)P(B') + P(A')P(B) = (1/2)(2/3) + (1/2)(1/3) = 2/6 + 1/6 = 3/6 = 1/2

NCERT/CBSEID: prob_exam_5

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. 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 hostlier?

Solution Strategy1.

E_1

: Hostlier,

E_2

: Day Scholar.

P(E_1)=0.6, P(E_2)=0.4

. 2. A: Grade A.

P(A|E_1)=0.3, P(A|E_2)=0.2

. 3. Bayes:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

4. Num =

0.6 \times 0.3 = 0.18

5. Denom =

0.18 + (0.4 \times 0.2) = 0.18 + 0.08 = 0.26

6. Ans:

0.18/0.26 = 18/26 = 9/13

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Probability .

The mathematics of uncertainty. From coin tosses to Bayes' prediction models.

Random Experiments & Events

Key Terms

Random Experiment:

An experiment whose outcome cannot be predicted with certainty (e.g., Tossing a coin).

Sample Space (S):

The set of all possible outcomes of a random experiment. Ex: Coin toss

S = \{H, T\}

Event (E):

Any subset of the sample space S. Ex: Getting a Head

E = \{H\}

Mutually Exclusive:

Events that cannot happen simultaneously.

A \cap B = \phi

Exhaustive Events:

Set of events whose union is the sample space.

E_1 \cup E_2 \cup ... = S

Types of Events

Impossible Event

P(\phi) = 0

Probability is 0. Ex: Rolling a 7 on a die.

Sure Event

P(S) = 1

Probability is 1. Ex: Rolling number < 7 on a die.

Simple Event

E = \{w_i\}

Has only one sample point.

Compound Event

Has more than one sample point.

Algebra of Probability

Understanding relationships between events using Set Theory (Union, Intersection, Complement).

Visualizing Probability

Formula & Concept

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Addition Theorem

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Example:

P(A)=0.6, P(B)=0.4, P(A \cap B)=0.2

, find

P(A \cup B)

P(A \cup B) = 0.6 + 0.4 - 0.2 = 0.8

Axiomatic Definition

Probability $P(E)$ is a real-valued function satisfying:

$0 \le P(E) \le 1$
$P(S) = 1$
If $A, B$ are mutually exclusive, $P(A \cup B) = P(A) + P(B)$

Conditional Probability & Bayes

Conditional Probability

Probability of event A happening given that event B has already happened.

Formula

P(A|B) = \frac{P(A \cap B)}{P(B)}, \text{ provided } P(B) \neq 0

Multiplication Thm

P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)

Properties

$P(S|F) = P(F|F) = 1$
$P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$
$P(E'|F) = 1 - P(E|F)$

Total Probability & Bayes Theorem

Law of Total Probability

Total probability of A occurring is sum of all paths leading to A:

P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)

Bayes Theorem (Reverse)

Given A happened, chance it came from path E1?

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

Example: Total Probability

Bag I: 3 Red, 4 Black. Bag II: 5 Red, 6 Black. One bag is selected at random and a ball is drawn. Find prob it is Red.

P(R) = P(E_1)P(R|E_1) + P(E_2)P(R|E_2)

= \frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}

= \frac{3}{14} + \frac{5}{22} = \frac{33+35}{154} = \frac{68}{154} = \frac{34}{77}

Example: Bayes Theorem

Bag I: 3 Red, 4 Black. Bag II: 5 Red, 6 Black. One ball drawn from selected bag is Red. Find prob it is from Bag II.

Identify the reverse probability question: "Given Outcome (Red), find Source (Bag II)". This is a classic Bayes signature.

1. Events:

E_1

(Bag I),

E_2

(Bag II),

A

(Red Ball) 2. Priors:

P(E_1)=P(E_2)=1/2

3. Conditionals:

P(A|E_1)=3/7

(Red from Bag I),

P(A|E_2)=5/11

(Red from Bag II) 4. Bayes Formula:

P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

5. Calc:

\frac{1/2 \cdot 5/11}{1/2 \cdot 3/7 + 1/2 \cdot 5/11} = \frac{5/22}{34/77} = \frac{35}{68}

✏️ Practice Zone

Problems

A coin is tossed three times. What is the probability of getting at least two heads?

✓ Show Solution

1. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8

2. E (at least 2 H) = {HHH, HHT, HTH, THH}, n(E) = 4

P(E) = 4/8 = 1/2

EASY

If A and B are two events such that

P(A) = 1/4, P(B) = 1/2

and

P(A \cap B) = 1/8

, find

P( ext{not A and not B})

✓ Show Solution

P(A' \cap B') = P((A \cup B)')

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1/4 + 1/2 - 1/8 = (2+4-1)/8 = 5/8

P(A' \cap B') = 1 - 5/8 = 3/8

MEDIUM

A die is thrown twice and the sum of numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

✓ Show Solution

1. E (sum 6) = {(1,5), (2,4), (3,3), (4,2), (5,1)}. n(E)=5.

2. F (4 at least once) = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}.

E \cap F = \{(2,4), (4,2)\}

n(E \cap F) = 2

P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{2/36}{5/36} = 2/5

MEDIUM

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

✓ Show Solution

P(B_1) = 26/52 = 1/2

P(B_2|B_1) = 25/51

(since one black card is gone)

P(B_1 \cap B_2) = P(B_1)P(B_2|B_1) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}

EASY

✓ Show Solution

P(M)=0.4, P(B)=0.25, P(M \cap B)=0.15

2. (i)

P(B|M) = \frac{P(B \cap M)}{P(M)} = \frac{0.15}{0.40} = \frac{15}{40} = 3/8

3. (ii)

P(M|B) = \frac{P(M \cap B)}{P(B)} = \frac{0.15}{0.25} = \frac{15}{25} = 3/5

MEDIUM

A die is tossed thrice. Find the probability of getting an odd number at least once.

✓ Show Solution

1. P(Odd) in one toss = 3/6 = 1/2. P(Even) = 1/2.

2. P(At least one Odd) = 1 - P(None Odd) = 1 - P(All Even)

3. Since tosses independent: P(All Even) =

P(E)P(E)P(E) = (1/2)^3 = 1/8

4. Ans:

1 - 1/8 = 7/8

MEDIUM

CBSE Exam Corner

Master these questions to ace your board exams.

CBSE SampleID: prob_exam_1

Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

Solution Strategy1. Let K, K, A be events. 2. P(K1) = 4/52 3. P(K2|K1) = 3/51 4. P(A|K1K2) = 4/50 5. Total Prob =

\frac{4}{52} \times \frac{3}{51} \times \frac{4}{50} = \frac{1}{13} \times \frac{1}{17} \times \frac{2}{25} = \frac{2}{5525}

CBSE 2023ID: prob_exam_2

In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective type. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

Solution Strategy1. Let

E_1, E_2, E_3

be events of Machine A, B, C.

P(E_1)=0.25, P(E_2)=0.35, P(E_3)=0.40

. 2. Let D be defective.

P(D|E_1)=0.05, P(D|E_2)=0.04, P(D|E_3)=0.02

. 3. Bayes:

P(E_2|D) = \frac{P(E_2)P(D|E_2)}{\sum P(E_i)P(D|E_i)}

4. Numerator:

0.35 \times 0.04 = 0.0140

5. Denom:

0.25(0.05) + 0.35(0.04) + 0.40(0.02) = 0.0125 + 0.0140 + 0.0080 = 0.0345

6. Ans:

\frac{0.0140}{0.0345} = \frac{140}{345} = 28/69

CBSE ModelID: prob_exam_3

A and B toss a coin alternately till one of them tosses a head and wins the game. If A starts the game, find their respective probabilities of winning.

Solution Strategy1. P(H) = 1/2, P(T) = 1/2. 2. A wins in 1st (H), 3rd (TTH), 5th (TTTTH)... toss. 3. P(A wins) =

1/2 + (1/2)^3 + (1/2)^5 + ...

4. This is a GP with

a=1/2, r=1/4

. 5. Sum =

\frac{a}{1-r} = \frac{1/2}{1-1/4} = \frac{1/2}{3/4} = 2/3

. 6. P(B wins) =

1 - P(A) = 1 - 2/3 = 1/3

ImportantID: prob_exam_4

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 solves the problem.

Solution Strategy1.

P(A)=1/2, P(B)=1/3

P(A')=1/2, P(B')=2/3

. 2. (i) Prob problem solved =

P(A \cup B) = 1 - P(A' \cap B') = 1 - P(A')P(B') = 1 - (1/2)(2/3) = 1 - 1/3 = 2/3

. 3. (ii) Exactly one =

P(A)P(B') + P(A')P(B) = (1/2)(2/3) + (1/2)(1/3) = 2/6 + 1/6 = 3/6 = 1/2

NCERT/CBSEID: prob_exam_5

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. 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 hostlier?

Solution Strategy1.

E_1

: Hostlier,

E_2

: Day Scholar.

P(E_1)=0.6, P(E_2)=0.4

. 2. A: Grade A.

P(A|E_1)=0.3, P(A|E_2)=0.2

. 3. Bayes:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

4. Num =

0.6 \times 0.3 = 0.18

5. Denom =

0.18 + (0.4 \times 0.2) = 0.18 + 0.08 = 0.26

6. Ans:

0.18/0.26 = 18/26 = 9/13