We already know that probability calculates the chance of happening any event.

The formula for calculation of probability is:

Probability\ =\frac{No.\ of\ Favorable\ Outcomes}{Total\ Number\ of\ Possible\ Outcomes}\\\ \\

Now there are certain axioms/rules which govern the calculation of probability.

In this post we will try to understand these axioms and solve questions related to the rules.

**Rules of Probability**

Suppose we performed one experiment and its sample size is given as:

Sample Size S = {W1, W2, W3, W4, W5, W6}

Here W1, W2 . . . . . are the sample elements

**Probability Axiom 01**

The probability of any event cannot be negative.

Its value lies between 0 and 1.

0\leqq \ P( Wi) \ \ \leqq 1

**Probability Axiom 02**

The probability of sample space is equal to 1.

This is because all the possible element is present in sample space which means that there is 100% chance that event will occur.**P (S) = 1 **

**Probability Axiom 03**

If elements are disjoint then the probability of individual element is equal to 1**P(W1) + P(W2) + P(W3) + P(W4) + P(W5) + P(W6) = 1**

**Probability Axiom 04**

If the elements are disjoint and independent, then the probability of event can be calculated by adding the probability of individual element**P(A) = ∑ P(ωi )**

Here

A = Event**ω** = element of sample space

**Other Important Probability Formulas**

**(1) Probability of Event A or B**

If there are two events A & B, then the probability of happening of event A or B is given by formula:

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

Where

P(A) = Probability of Event A

P(B) = Probability of Event B

P( A\cap B) = Probability of intersection of event A & B

**(2) Probability of Event ” Not A”**

This means that we have to find probability of elements outside event A.

For this operation we have simple formula:**P(A’) = 1 – P(A)**

**Questions on Probability Formulas**

**(01) Three Pens are manufactured. The pens can be classified into two categories Good (G) and Defective (D).**

Find the probability of following events:

(i) Event A : getting exactly one defective pen

(ii) Event B: getting atleast two defective pens

**Solution**

Let us first find the sample size of above experiment

Sample Size S = {GGD, GDG, GGG, GDD, DGD, DDG, DGG, DDD}

(i) Event A : getting exactly one defective pen

Event A : {GGD, GDG, DGG}

From the above illustration you can observe that:

P(GGD) = 1/8

P(GDG) = 1/8

P(DGG) = 1/8

Now using the axiom 4 we know that probability of event can be calculated by adding probability of individual element (if the elements are independent)

So probability of event A = 1/8+1/8+/8 ⟹ **3/8**

(ii) Event B : Getting atleast two defective pens

Event B : {GDD, DGD, DDG, DDD}

From the above illustration you can observe that:

P (GDD) = 1/8

P (DGD) = 1/8

P (DDG) = 1/8

P (DDD) = 1/8

Again sing Axiom 4, the probability of event B = 1/8+1/8+/8 +1/8 = 1/2

**(02) Three coins are tossed simultaneously. Two events are provided as follows:**

Event A = {HHT, HTH, THH}

Event B = {HTH, THH, HHH}

find the probability of event ‘A or B’, i.e., P (A ∪ B)

**Solution**

The formula for P (A ∪ B) is:

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

Let us calculate the probability of P(A)

From the above illustration you can see that **P(A) = 3/8**

Similarly the probability of event B, **P(B) = 3/8**

Now we have to find probability of P( A\cap B)

For this let us find the common element between Event A & B

Event A = {HHT, HTH, THH}

Event B = {HTH, THH, HHH}

You can see that there is no common element between the two events.

Hence P( A\cap B) = 0

Putting all the values in equation (1), we get

P\ ( A\ \cup \ B) \ =\ P( A) \ +P( B) \ -P( A\cap B)\\\ \\ P\ ( A\ \cup \ B) \ =\ \frac{3}{8} +\frac{3}{8} \ -\ 0\\\ \\ P\ ( A\ \cup \ B) \ =\ \frac{6}{8} \\\ \\

**Hence 6/8 is the required answer**

**(03) A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag**.

Calculate the probability that it will be

(i) red

(ii) yellow

(iii) blue

(iv) not blue

(v) either red or blue

**Solution**

For this Question the sample size will be:

S = {R1, R2, R3, R4, B1, B2, B3, Y1, Y2}**(i) Event A = Ball is red**

Probability Formula = (Number of favorable Outcome/ Total Possible Outcome)

Favorable Outcomes for Event A = {R1, R2, R3}

Probability of Event A, P(A) = 4/9**(ii) Event B = Ball is Yellow**

Favorable Outcome for Event B = {Y1, Y2}

Probability of Event B, P(B) = 2/9**(iii) Event C = Ball is Blue**

Favorable Outcome for Event C = {B1, B2, B3}

Probability of Event C, P(C) = 3/9

**(iv) Not Blue**

Here we have to find value of P(C’)

we know that P(C’) = 1 – P(C)

⟹ P(C’) = 1 – 3/9

⟹ P(C’) = 6/9

Hence 6/9 is the required probability

**(v) Either Red or Blue**

we have to calculate P(AUC)

P\ ( A\ \cup \ C) \ =\ P( A) \ +P( C) \ -P( A\cap C)\\\ \\ P\ ( A\ \cup \ C) \ =\ \frac{4}{9} +\frac{3}{9} \ -\ 0\\\ \\ P\ ( A\ \cup \ C) \ =\ \frac{7}{9}

**(04) One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be**:

(i) a diamond

(ii) not an ace

(iii) a black card (i.e., a club or, a spade)

(iv) not a diamond

(v) not a black card.

Solution

(i) Event A = Card is Diamond

Probability = Number of Favorable Outcome/ Total Possible Outcome

P(A) = 13/52

(ii) Event B = Card is not an Ace

Total Non Ace Cards = 52 – 4 = 48

P(B) = 48/52

(iii) Event C = A black Card

Total Black Cards = 26

P (C)= 26/52

(iv) Not a Diamond

we know P(A) = 13/52

So P(A’) = 1- 13/52 ⟹ 39/52

(v) Not a Black Card

P(C’) = 1 – P(C)

P(C’) = 1 – 26/52

P(C’) = 26/52