# Probability Theory: Best Answers 007EH

Welcome to **Lecture Note Nine – Probability theory,**

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# Probability Theory The Logic of Science

The word ‘Probability’ means the chance of occurring of a particular event. It is generally possible to predict the future of an event quantitatively with a certain probability of being correct. The probability is used in such cases where the outcome of the trial is uncertain.

## Probability Definition:

The probability of happening of an event A, denoted by P(A), is defined as

Thus, if an event can happen in m ways and fails to occur in n ways and m+n ways is equally likely to occur then the probability of happening of the event A is given by

And the probability of non-happening of A is

**Note:**

- The probability of an event which is certain to occur is one.
- The probability of an event which is impossible to zero.
- If the probability of happening of an event P(A) and that of not happening is P(A), then

P(A)+ P(A) = 1, 0 ≤ P(A) ≤ 1,0≤ P(A)≤1.

## Important Terms related to Probability:

**1. Trial and Event:** The performance of an experiment is called a trial, and the set of its outcomes is termed an event.

**Example :** Tossing a coin and getting head is a trial. Then the event is {HT, TH, HH}

**2. Random Experiment:** It is an experiment in which all the possible outcomes of the experiment are known in advance. But the exact outcomes of any specific performance are not known in advance.

**Example:**

- Tossing a Coin
- Rolling a die
- Drawing a card from a pack of 52 cards.
- Drawing a ball from a bag.

**3. Outcome:** The result of a random experiment is called an Outcome.

**Example:** 1. Tossing a coin is an experiment and getting head is called an outcome.

2. Rolling a die and getting 6 is an outcome.

**4. Sample Space:** The set of all possible outcomes of an experiment is called sample space and is denoted by S.

**Example:** When a die is thrown, sample space is S = {1, 2, 3, 4, 5, 6}

It consists of six outcomes 1, 2, 3, 4, 5, 6

#### Note1: If a die is rolled n times the total number of outcomes will be 6^{n}.

#### Note2: If 1 die rolled n times then n die rolled 1 time.

**5. Complement of Event:** The set of all outcomes which are in sample space but not an event is called the complement of an event.

**6. Impossible Events:** An event which will never be happened.

**Example1:** Tossing double-headed coins and getting tails in an impossible event.

**Example2:** Rolling a die and getting number > 10 in an impossible outcome.

P (impossible outcome) =0

**7. Sure Outcome/Certain Outcome:** An Outcome which will definitely be happen

**Example1:** Tossing double-headed coins and getting heads only.

**Example2:** Rolling a die and getting number < 6

P (sure outcome) = 1

{1, 2, 3, 4, 5 6} is called sure event

P (sure outcome) = 1

**8. Possible Outcome:** An outcome which is possible to occur is called Possible Outcome.

**Example1:** Tossing a fair coin and getting a head on it.

**Example2:** Rolling a die and getting an odd number.

**9. Equally Likely Events:** Events are said to be equally likely if one of them cannot be expected to occur in preference to others. In other words, it means each outcome is as likely to occur as any other outcome.

**Example:** When a die is thrown, all the six faces, i.e., 1, 2, 3, 4, 5 and 6 are equally likely to occur.

**10. Mutually Exclusive or Disjoint Events:** Events are called mutually exclusive if they cannot occur simultaneously.

**Example:** Suppose a card is drawn from a pack of cards, then the events getting a jack and getting a king are mutually exclusive because they cannot occur simultaneously.

**11. Exhaustive Events:** The total number of all possible outcomes of an experiment is called exhaustive events.

**Example:** In the tossing of a coin, either head or tail may turn up. Therefore, there are two possible outcomes. Hence, there are two exhaustive events in tossing a coin.

**12. Independent Events:** Events A and B are said to be independent if the occurrence of any one event does not affect the occurrence of any other event.

P (A ∩ B) = P (A) P (B).

**Example:** A coin is tossed thrice, and all 8 outcomes are equally likely

A: “The first throw results in heads.”

B: “The last throw results in Tails.”

Prove that event A and B are independent.

**Solution:**

**13. Dependent Event:** Events are said to be dependent if occurrence of one affect the occurrence of other events.

# Addition Theorem

**Theorem1:** If A and B are two mutually exclusive events, then

P(A ∪B)=P(A)+P(B)

**Proof:** Let the n=total number of exhaustive cases

n_{1}= number of cases favorable to A.

n_{2}= number of cases favorable to B.

Now, we have A and B two mutually exclusive events. Therefore, n_{1}+n_{2} is the number of cases favorable to A or B.

**Example:** Two dice are tossed once. Find the probability of getting an even number on first dice or a total of 8.

**Solution:** An even number can be got on a die in 3 ways because any one of 2, 4, 6, can come. The other die can have any number. This can happen in 6 ways.

∴ P (an even number on Ist die) =

A total of 8 can be obtained in the following cases:

{(2,6),(3,5),(4,4),(5,3),(6,2)}

∴ P (a total of 8) =

∴ Total Probability =

**Theorem2:** If A and B are two events that are not mutually exclusive, then

P(A ∪B)=P(A)+P(B)- P (A∩B).

**Proof:** Let n = total number of exhaustive cases

n_{1}=number of cases favorable to A

n_{2}= number of cases favorable to B

n_{3}= number of cases favorable to both A and B

But A and B are not mutually exclusive. Therefore, A and B can occur simultaneously. So,n_{1}+n_{2}-n_{3} is the number of cases favorable to A or B.

Therefore, P(A ∪B)=

But we have, P(A)=, P(B) =and P (A∩B)=

Hence, P(A ∪B)=P(A)+P(B)- P (A∩B).

**Example1:** Two dice are tossed once. Find the probability of getting an even number on first dice or a total of 8.

**Solution:** P(even number on Ist die or a total of 8) = P (even number on Ist die)+P (total of 8)= P(even number on Ist die and a total of 8)

∴ Now, P(even number on Ist die)=

Ordered Pairs showing a total of 8 = {(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)} = 5

∴ Probability; P(total of 8) =

P(even number on Ist die and total of 8) =

∴ Required Probability =

**Example2:** Two dice are thrown. The events A, B, C, D, E, F

A = getting even number on first die.

B= getting an odd number on the first die.

C = getting a sum of the number on dice ≤ 5

D = getting a sum of the number on dice > 5 but less than 10.

E = getting sum of the number on dice ≥ 10.

F = getting odd number on one of the dice.

**Show the following:**

1. A, B are a mutually exclusive event and Exhaustive Event.

2. A, C are not mutually exclusive.

3. C, D are a mutually exclusive event but not Exhaustive Event.

4. C, D, E are a mutually exclusive and exhaustive event.

5. A’∩B’ are a mutually exclusive and exhaustive event.

6. A, B, F are not a mutually exclusive event.

**Solution:**

**A:** (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

**B:** (1,1), (1,2),(1,3),(1,4),(1,5),(1,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

**C:** (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)

**D:** (1,5),(1,6),(2,4),(2,5),(2,6)

(3,3),(3,4),(3,5),(3,6)

(4,2),(4,3),(4,4),(4,5)

(5,1),(5,2),(5,3),(5,4)

(6,1),(6,2),(6,3)

**E:** (4,6),(5,5),(5,6),(6,5),(6,6),(6,4)

**F:** (1,2),(1,4),(1,6)

(2,1),(2,3),(2,5)

(3,2),(3,4),(3,6)

(4,1),(4,3),(4,5)

(5,2),(5,4),(5,6)

(6,1),(6,3),(6,5)

**1.** (A∩B) =∅ and (A∪B)=S

A, B are a mutually exclusive and exhaustive event.

**2.** (A∩C) are not mutually exclusive

(2,1),(2,3),(4,1)≠ ∅

**3.** C∩D are a mutually exclusive but not exhaustive event.

C∩D=∅ C∪ D≠S

**4.** C∩D=∅,D∩E=∅, C∩E=∅ are mutually exclusive and exhaustive event.

**5.** A’∩B’ =(A∪B)’ are a mutually exclusive and exhaustive event.

**6.** (A∩B) =∅ are a mutually exclusive

A, B, F are not mutually exclusive events.

# Multiplication Theorem

**Theorem:** If A and B are two independent events, then the probability that both will occur is equal to the product of their individual probabilities.

P(A∩B)=P(A)xP(B)

**Proof:** Let event

A can happen is n_{1}ways of which p are successful

B can happen is n_{2}ways of which q are successful

Now, combine the successful event of A with successful event of B.

Thus, the total number of successful cases = p x q

We have, total number of cases = n_{1} x n_{2}.

Therefore, from definition of probability

P (A and B) =P(A∩B)=

We have P(A) =,P(B)=

So, P(A∩B)=P(A)xP(B)

If, there are three independent events A, B and C, then

P(A∩B∩C)=P((A∩B)∩C)= P(A∩B)xP(C)

=P(A) x P(B) x P(C).

In general, if there are n independent events, then

**Example:** A bag contains 5 green and 7 red balls. Two balls are drawn. Find the probability that one is green and the other is red.

**Solution:** P(A) =P(a green ball) =

P(B) =P(a red ball) =

By Multiplication Theorem

P(A) and P(B) = P(A) x P(B) =

# Conditional Probability

**Theorem:** If A and B are two dependent events then the probability of occurrence of A given that B has already occurred and is denoted by P(A/B) is given by

Similarly, the probability of occurrence of B given that A has already occurred is given by

**Proof:** Let S be the sample space.

Then, we have

Interchange A and B in equation (i), we get

**Example:** Find the probability of drawing a heart on each of two consecutive draws from well shuffled-packs of cards if the card is not replaced after the draw.

**Solution:** Let event A is a heart on the first draw, and event B is a heart on the second draw.

When we get a heart on the first draw, the second draw has 51 outcomes and 12 are favorable.

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