Probability Axioms and Formulas for Grade 11 Math

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.

Probability axioms and rules for class 11 NCERT/CBSE Math

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

Important formulas of Probability for Grade 11 CBSE/NCERT Maths

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 )

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) \\\ \\

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

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}

Class 11 questions on probability with solutions

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}

Important concepts and formulas for probability for Class 11 CBSE and NCERT  Math

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)

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)

Concepts of probability for high class math

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

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.

(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

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