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(i) A spade

Total numbers of cards are 52

Total number of spade cards = 13

Probability of getting spade is = Total number of spade cards/Total number of cards

= 13/52

= 1/4

∴ Probability of getting a spade is 1/4

(ii) A black card

Total numbers of cards are 52

Cards of spades and clubs are black cards.

Number of spades = 13

Number of clubs = 13

Total number of black card out of 52 cards = 13 + 13 = 26

Probability of getting black cards is = Total number of black cards/Total number of cards

= 26/52

= 1/2

∴ Probability of getting a black card is 1/2

(iii) The seven of clubs

Total numbers of cards are 52

Total number of the seven of clubs cards = 1

Probability of getting the seven of clubs cards is = Total number of the seven of club cards/ Total numbers of cards

= 1/52

∴ Probability of the seven of club card is 1/52

(iv) Jack

Total numbers of cards are 52

Total number of jack cards = 4

Probability of getting jack cards is = Total number of jack cards/ Total numbers of cards

= 4/52

= 1/13

∴ Probability of the jack card is 1/13

(v) The ace of spades

Total numbers of cards are 52

Total number of the ace of spades cards = 1

Probability of getting ace of spade cards is = Total number of ace of spade cards/ Total numbers of cards

= 1/52

∴ Probability of the ace of spade card is 1/52

(vi) A queen

Total numbers of cards are 52

Total number of queen cards = 4

Probability of getting queen cards is = Total number of queen cards/Total numbers of cards

= 4/52

= 1/13

∴ Probability of a queen card is 1/13

(vii) A heart

Total numbers of cards are 52

Total number of heart cards = 13

Probability of getting queen cards is = Total number of heart cards/Total numbers of cards

= 13/52

= 1/4

∴ Probability of a heart card is 1/4

(viii) A red card

Total numbers of cards are 52

Total number of red cards = 13+13 = 26

Probability of getting queen cards is = Total number of red cards/Total numbers of cards

= 26/52

= 1/2

∴ Probability of a red card is 1/2.

(i) a black king Total numbers of cards are 52 

Number of black king cards = 2 

Probability of getting black king cards is = \(\frac{Total\,number\,of\,blackking\,cards}{Total\,number\,of\,cards}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Therefore Probability of getting black king cards is = \(\frac{1}{26}\) 

(ii) either a black card or a king 

Total numbers of cards are 52 

Number of either a black card or a king = 28 

Probability of getting either a black card or a king is = \(\frac{Total\,number\,of\,blackking\,cards}{Total\,number\,of\,cards}\) =\(\frac{28}{52}\) = \(\frac{7}{13}\)

Therefore Probability of getting either a black card or a king is = \(\frac{7}{13}\)

(iii) black and a king 

Total numbers of cards are 52 

Number of black and a king = 2 

Probability of getting black and a king is = \(\frac{Total\,number\,of\,blackking\,cards}{Total\,number\,of\,cards}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\) 

Therefore Probability of getting black and a king is = \(\frac{1}{26}\)

(iv) a jack, queen or a king 

Total numbers of cards are 52 

Number of a jack, queen or a king = 12 

Probability of getting a jack, queen or a king is = \(\frac{Total\,number\,of\,blackking\,cards}{Total\,number\,of\,cards}\) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

Therefore Probability of getting a jack, queen or a king is = \(\frac{3}{13}\)

(v) neither a heart nor a king 

Total numbers of cards are 52 

Total number of heart cards = 13

Probability of getting a heart is = \(\frac{Total\,number\,of\,blackking\,cards}{Total\,number\,of\,cards}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

Total probability of getting a heart and a king = \(\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}\)

Therefore probability of getting neither a heart nor a king = \(1-\frac{4}{13}=\frac{9}{13}\)

(i) a spade

Total numbers of cards are 52

Total number of spade cards = 13

Probability of getting spade is = \(\frac{Total\,number\,of\,spade\,cards}{Total\,number\,of\,cards}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\)

(ii) a black card

Total numbers of cards are 52

Cards of spades and clubs are black cards.

Number of spades = 13

Number of clubs = 13

Therefore, total number of black card out of 52 cards = 13 + 13 = 26

Probability of getting black cards is = \(\frac{Total\,number\,of\,black\,cards}{Total\,number\,of\,cards}\) = \(\frac{26}{52}=\frac{1}{2}\) 

(iii) the seven of clubs

Total numbers of cards are 52

Number of the seven of clubs cards = 1

Probability of getting the seven of clubs cards is = \(\frac{Total\,number\,of\,the\,seven\,of\,clubs\,cards}{Total\,number\,of\,cards}\)

= \(\frac{1}{52}\)

(iv) jack

Total numbers of cards are 52

Number of jack cards = 4

Probability of getting jack cards is = \(\frac{Total\,number\,of\,jack\,cards}{Total\,number\,of\,cards}\) = \(\frac{4}{52}=\frac{1}{13}\)

(v) the ace of spades

Total numbers of cards are 52

Number of the ace of spades cards = 1

Probability of getting ace of spades cards is = \(\frac{Total\,number\,of\,ace\,of\,spade\,cards}{Total\,number\,of\,cards}\) 

= \(\frac{1}{52}=\frac{1}{52}\)

Total number of balls = 3

(i) 1/3

(ii) 1/3

(iii) 1/3

Given: pack of 52 cards 

Formula: P(E) = \(\frac{favourable\,outcomes}{total\,possible\,outcomes}\)

since a card is drawn from a pack of 52 cards, therefore number of elementary events in the sample space is 

n(S)= ⁵²C₁ = 52 

(i) let E be the event of drawing a black king 

n(E)=²C₁ =2 (there are two black kings one of spade and other of club)

P(E) = \(\frac{n(E)}{n(S)} \)

 P(E) = \(\frac{2}{52}\) = \(\frac{1}{26}\) 

(ii) let E be the event of drawing a black card or a king

n(E) = ²⁶C₁+⁴C₁- ²C₁= 28 

we are subtracting 2 from total because there are two black king which are already counted and to avoid the error of considering it twice

P(E) = \(\frac{n(E)}{n(S)} \)

 P(E) = \(\frac{28}{52}\) = \(\frac{7}{13}\) 

(iii) let E be the event of drawing a black card and a king 

n(E)=²C₁ =2 (there are two black kings one of spade and other of club)

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{5}{52}\) = \(\frac{1}{26}\)  

(iv) let E be the event of drawing a jack, queen or king 

n(E)=⁴C₁+ ⁴C₁+ ⁴C₁ = 12

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{12}{52}\) = \(\frac{3}{13}\)  

(v) let E be the event of drawing neither a heart nor a king now consider E’ as the event that either a heart or king appears 

n(E’) = ⁶C₁+ ⁴C₁ - 1 = 16 (there is a heart king so it is deducted)

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{16}{52}\) = \(\frac{4}{13}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{4}{13}\) = \(\frac{9}{13}\) 

(vi) let E be the event of drawing a spade or king 

n(E)= ¹³C₁ + ⁴C₁ - 1 = 16

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{16}{52}\) = \(\frac{4}{13}\)  

(vii) let E be the event of drawing neither an ace nor a king now consider E’ as the event that either an ace or king appears

n(E’) = ⁴C₁+ ⁴C₁ = 8

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{8}{52}\) = \(\frac{2}{13}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{2}{13}\) = \(\frac{11}{13}\) 

(viii) let E be the event of drawing a diamond card 

n(E)= ¹³C₁ = 13

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{13}{52}\) = \(\frac{1}{13}\)   

(ix) let E be the event of drawing not a diamond card now consider E’ as the event that diamond card appears 

n(E’) = ¹³C₁ =13

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{13}{52}\) = \(\frac{1}{4}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) 

(x) let E be the event of drawing a black card 

n(E)= ²⁶C₁= 26 (spades and clubs)

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{26}{52}\) = \(\frac{1}{2}\)   

(xi) let E be the event of drawing not an ace now consider E’ as the event that ace card appears

n(E’) = ⁴C₁= 4

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{4}{52}\) = \(\frac{1}{13}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{1}{13}\) = \(\frac{12}{13}\)  

(xii) let E be the event of not drawing a black card 

n(E) = ²⁶C₁= 26 (red cards of hearts and diamonds)

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{26}{52}\) = \(\frac{1}{2}\) 

Total number of outcomes, n(S) = 52 

(i) n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(ii) n(E) = 26 + 2 = 28

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{28}{52}\) = \(\frac{7}{13}\)

(iii) n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(iv) n(E) = 12

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(v) n(E) = 36

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{36}{52}\) = \(\frac{9}{13}\)

(vi) n(E) = 16

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(vii) n(E) = 44

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{44}{52}\)

(viii) n(E) = 24

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{24}{52}\) = \(\frac{6}{13}\)

(ix) n(E) = 48

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{48}{52}\) = \(\frac{12}{13}\)

(x) n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(xi) n(E) = 13

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\)

(xii) n(E) = 26

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{26}{52}\) = \(\frac{1}{2}\)

(xiii) n(E) = 1

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{52}\)

(xiv) n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(xv) n(E) = 1

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{52}\)

 (xvi) n(E) = 4

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(xvii) n(E) = 13

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\)

(xviii) n(E) = 26

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{26}{52}\) = \(\frac{1}{2}\)

(xix) n(E) = 44

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{44}{52}\) = \(\frac{11}{13}\)

Given: Three dice are rolled over.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

So, we now have to determine the probability of getting the same number on all the three dice

Total number of possible outcomes is 6³ = 216

n (S) = 216

Let E be the event of getting same number on all the three dice

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

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 216

= 1/36

Given: three dice are rolled

Formula: P(E) = \(\frac{favourable\,outcomes}{total\,possible\,outcomes}\) 

so, we have to determine the probability of getting the same number on all the three dice total number of possible outcomes are 6² = 36

therefore n(S)=36 

let E be the event of getting same number on all the three dice 

E= {(5,6) (6,5) (6,6)}

n(E)= 3

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{3}{36}\) = \(\frac{1}{12}\) 

Given: pack of 52 cards from which 4 are dropped

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

we have to find the probability that all the face cards of same suits are drawn total possible outcomes are 

n(S) = ⁵²C₄ 

let E be the event that all the cards drawn are face cards of same suit 

n(E)= 4 × ⁴C₄ = 4

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{4}{270725}\) 

Given: Two unbiased dice are thrown.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

So, we now have to determine the probability of getting the sum of digits on dice greater than 10

Total number of possible outcomes is 6²=36

n (S) = 36

Let E be the event of getting same number on all the three dice

E = {(5,6) (6,5) (6,6)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(i) A black king

Total number of cards are 52

Number of black king cards = 2

Probability of getting black king cards is = Total number of black king cards/Total number of cards

= 2/52

= 1/26

∴ Probability of getting black king cards is 1/26

(ii) Either a black card or a king

Total number of cards are 52

Number of either a black card or a king = 28

Probability of getting either a black card or a king is = Total number of either black or king card/Total number of cards

= 28/52

= 7/13

∴ Probability of getting either a black card or a king is 7/13

(iii) Black and a king

Total number of cards are 52

Number of black and a king = 2

Probability of getting black and a king is = Total number of black and king card/Total number of cards

= 2/52

= 1/26

∴ Probability of getting black and a king is 1/26

(iv) a jack, queen or a king

Total number of cards are 52

Number of a jack, queen or a king = 12

Probability of getting a jack, queen or a king is = Total number of jack, queen or king card/Total number of cards

= 12/52

= 3/13

∴ Probability of getting a jack, queen or a king is 3/13

(v) Neither a heart nor a king

Total numbers of cards are 52

Total number of heart cards = 13

Probability of getting a heart is = Total number of hearts/Total number of cards

= 13/52

= 1/4

Total number of king cards = 4

Probability of getting a king is = Total number of king card/Total number of cards

= 4/52

= 1/13

Total probability of getting a heart and a king = 13/52 + 4/52 – 1/52

= (13+4-1)/52

= 16/52

= 4/13

∴ Probability of getting neither a heart nor a king = 1 – 4/13 = (13-4)/13 = 9/13

Given: A deck of 52 cards.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

We now have to find the probability that all the face cards of same suits are drawn.

Total possible outcomes are

n (S) = ⁵²C₄

Let E be the event that all the cards drawn are face cards of same suit.

n (E) = 4 × ⁴C₄ = 4

P (E) = n (E) / n (S)

= 4 / 270725

Given: Pack of 52 cards.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

We know that, a card is drawn from a pack of 52 cards, so number of elementary events in the sample space is

n (S) = ⁵²C₁ = 52

(i) Let E be the event of drawing a black king

n (E) = ²C₁ = 2 (there are two black kings one of spade and other of club)

P (E) = n (E) / n (S)

= 2 / 52

= 1/26

(ii) Let E be the event of drawing a black card or a king

n (E) = ²⁶C₁ +⁴C₁ – ²C₁= 28

[We are subtracting 2 from total because there are two black king which are already counted and to avoid the error of considering it twice.]

P (E) = n (E) / n (S)

= 28 / 52

= 7/13

(iii) Let E be the event of drawing a black card and a king

n (E) = ²C₁ = 2 (there are two black kings one of spade and other of club)

P (E) = n (E) / n (S)

= 2 / 52

= 1/26

(iv) Let E be the event of drawing a jack, queen or king

n (E) = ⁴C₁ + ⁴C₁ + ⁴C₁ = 12

P (E) = n (E) / n (S)

= 12 / 52

= 3/13

(v) Let E be the event of drawing neither a heart nor a king

Now let us consider E′ as the event that either a heart or king appears

n (E′) = ⁶C₁ + ⁴C₁ - 1 = 16 (there is a heart king so it is deducted)

P (E′) = n (E′) / n (S)

= 16 / 52

= 4/13

So, P (E) = 1 – P (E′)

= 1 – 4/13

= 9/13

(vi) Let E be the event of drawing a spade or king

n (E) = ¹³C₁ + ⁴C₁ - 1 = 16

P (E) = n (E) / n (S)

= 16 / 52

= 4/13

(vii) Let E be the event of drawing neither an ace nor a king

Now let us consider E′ as the event that either an ace or king appears

n(E′) = ⁴C₁ + ⁴C₁ = 8

P (E′) = n (E′) / n (S)

= 8 / 52

= 2/13

So, P (E) = 1 – P (E′)

= 1 – 2/13

= 11/13

(viii) Let E be the event of drawing a diamond card

n (E)=¹³C₁=13

P (E) = n (E) / n (S)

= 13 / 52

= 1/4

(ix) Let E be the event of drawing not a diamond card

Now let us consider E′ as the event that diamond card appears

n (E′) =¹³C₁=13

P (E′) = n (E′) / n (S)

= 13 / 52

= 1/4

So, P (E) = 1 – P (E′)

= 1 – 1/4

= 3/4

(x) Let E be the event of drawing a black card

n (E) =²⁶C₁ = 26 (spades and clubs)

P (E) = n (E) / n (S)

= 26 / 52

= 1/2

(xi) Let E be the event of drawing not an ace

Now let us consider E′ as the event that ace card appears

n (E′) = ⁴C₁ = 4

P (E′) = n (E′) / n (S)

= 4 / 52

= 1/13

So, P (E) = 1 – P (E′)

= 1 – 1/13

=12/13

(xii) Let E be the event of not drawing a black card

n (E) = ²⁶C₁ = 26 (red cards of hearts and diamonds)

P (E) = n (E) / n (S)

= 26 / 52

= 1/2

Given: A pack of 52 cards from which 4 are dropped.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

We now have to find the probability that the missing cards should be one from each suit

We know that, from well shuffled pack of cards, 4 cards missed out total possible outcomes are

n (S) = ⁵²C₄ = 270725

Let E be the event that four missing cards are from each suite

n (E) = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ = 13⁴

P (E) = n (E) / n (S)

= 13⁴ / 270725

= 2197/20825

N(S) = 36 

Let A : sum is 11 = {(6, 5) (5, 6)) 

⇒ n(A) = 2 

(a) P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{36}\)

= \(\frac{1}{18}\)

(b) Let B = Sum> 11 = {(66)),n(B) = 1 

P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{1}{36}\) 

(c) Let C = multiple of 2 on one die & multiple of 3 on the other 

{(2, 3), (2, 6), (4, 3) (4, 6) (6, 3) (6, 6)} n(c) 

= 6 P(c) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

S = {HHH, HHT, HTH, THH TTH, THT, HTT, TTT}

Correct option is (A) No. of favourable outcomes

\(\because\) P(E) \(=\frac{n(E)}{n(S)};\) where n(E) represents no. of favourable outcomes & n(S) represents total possible outcomes.

Hence, option (A) is correct.

A) No. of favourable outcomes

1) Find the probability of getting an even number when the dice is thrown once? 

2) What is the probability of getting an odd number when a dice is thrown once?

The possible outcomes : 

A, B and C.

Correct option is (C) 1/2

When a coin is tossed, possible outcomes are head and tail.

S = {H, T}

Total outcomes is n(S) = 2

Number of favourable outcomes of event of getting a head is n(E) = 1.

\(\therefore\) Probability of getting a head is P(E) \(=\frac{n(E)}{n(S)}=\frac12.\)

Correct option is  C) 1/2 

Let E denotes the event that student passed in first examination. 

And H be the event that student passed in second exam. 

Given, P(E) = 0.8 and P(H) = 0.7 

Also probability of passing atleast one exam i.e P(E or H) = 0.95 

Or, P(E∪H) = 0.95 

We have to find the probability of the event in which students pass both the examinations i.e. P(E∩H) 

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that: 

P(A∪B) = P(A) + P(B) – P(A∩B) 

∴ P(E∪H) = P(E) + P(H) – P(E∩H) 

⇒ P(E∩H) = P(E) + P(H) – P(E∪H) 

⇒ P(E∩H) = 0.7 + 0.8 – 0.95 

= 1.5 – 0.95 = 0.55 

∴ Probability of passing both the exams = P(E∩H) = 0.55

Correct option is (A) event

The each outcome of a random experiment is called primary event.

Correct option is  A) event

Total number of students = 60 = Number of possible outcomes. 

Number of students take tea = 32 

So number of students who don’t take tea 

= 60 – 32 = 28 

Number of favourable outcomes for choosing a boy who doesn’t take tea = 28 

∴ Its probability = \(\frac{28}{60}=\frac{7}{15}\)

An event is anything whose probability we want to measure, as getting an even number on throwing a dice.

An outcome is any way in which an event can happen.

Correct option is: A) even number and odd numbe

Sample space is the set of all possible outcomes of an experiment, generally denoted by S. 

Ex. When a dice is thrown, S = { 1, 2, 3, 4, 5, 6}

Equally likely events : 

Two events are said to be equally likely events if the probability of occurrence of those events in that experiment is equal.

Equally likely events: When two or more events have an equal chance of happening, then they are called equally likely.

Ex. In a throw of a dice, all the six faces (1, 2, 3, 4, 5, 6) are equally likely to occur.

The probability of a certain event is 1 and the probability of an impossible event is 0. Probability is never greater than 1 or less than 0.

Generally probability examples involve coins, dice and pack of cards. Here is a remainder of their outcomes. 

(i) A toss of a coin has two outcomes: head or tail. 

(ii) A throw of a dice has six outcomes: 1, 2, 3, 4, 5, 6. 

(iii) A pack of cards consists of 52 cards, divided into 4 suits, each suit containing 13 cards. 

Hearts (Red), Spades (Black), Diamonds (Red) and Clubs (Black). Each of the suits has 13 cards bearing the values 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King and Ace. The Jack, Queen and King are called ‘picture cards’. 

Hence there are 12 picture cards in all. 

Total number of possible outcomes = 52

Probability of an event = P(E) = \(\frac{Number\,of\,ways\,the\,event\,can\,happen}{Total\,number\,of\,possible\,outcomes}\)

Ex. P(drawing a picture card) = \(\frac{12}{52}\) = \(\frac{3}{13}\) ( \(\therefore\) There are 12 picture cards)

Let S be the sample space of a random experiment since S is a subset of itself, it also represents an event. Since all the outcomes of a random experiment belong to sample space S, the event S represents a sure event.

Events such as tossing a head or a tail with a coin, drawing a Queen or a Jack from a pack of cards, throwing an even or a odd number with a dice are all mutually exclusive events. Here, the occurrence of an event rules out the happening of all the other events in the same experiment, i.e., If we toss a coin, we can never get a head or a tail in the same toss.

Probability (head) =\(\frac{1}{2}\) , Probability (tail) =\(\frac{1}{2}\)

Also, Probability (head) + Probability (tail) = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

Such events are also called exhaustive events, because there are no other possibilities and their probabilities always add up to 1.

An example of events that are not mutually exclusive would be throwing a prime number or an odd number with a dice. There are two prime number 3 and 5 which are also odd numbers.

When two events are mutually exclusive, we can find the probability of either of them occurring by adding together the separate probabilities. 

Ex. The probability of throwing a 3 or a 5 with a dice is

P(3) + P(5) = \(\frac{1}{6}+\frac{1}{6}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\).

If we assume that S is the sample space of a random experiment then ϕ (null set) is a subset of S. 

As no outcome of the experiment belong to ϕ, the event ϕ can never occur when the experiment is performed. 

Hence ϕ is an impossible event.

Let S be the sample space associated with a random experiment and let E be an event. The event of non-occurrence of E is called the complementary event of E. 

It is represented by \(\bar{E}\) or E′. \(\bar{E}\) is also known as not E. 

Ex. If in a throw of a die, getting a prime number is a favourable event A, then the event of getting a number that is not prime is called the complementary event of A. 

If A = {2, 3, 5}, then \(\bar{A}\) = {1, 4, 6}

Complementary event: The complementary event of A = Event A not happening. Thus, an event and its complementary event are both mutually exclusive and exhaustive. Hence, Probability (event A not happening) = 1 – Probability (event A happening) Thus, P(\(A\)) + P(\(\bar A\)) = 1, where \(\bar A\) denotes the complementary event of A. 

Ex. Probability of drawing a blue ball from a bag of 4 blue, 6 red and 2 yellow balls

= \(\frac{4}{4+6+2}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)

\(\therefore\) P (not drawing a blue ball) = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

Answer is (A) 0.0001