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Total number of balls = 20

(i) Number of White Balls = 4

Total Balls = 20

P(white) = 4/20 = 1/5

(ii) No black means the we have to pick white, red, blue balls only

P(not black) = 13/20

(iii) neither black nor white means only white and red balls are left

P(neither white nor black) = 9/20

(i) Total number cards in a deck = 52 

Number of 4’s in a deck of cards = 4

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,in\,a\,deck}\)

∴ Probability of drawing a 4 numbered card from the deck of cards P(4)

= \(\frac{Numbe\,of\,4's\,in\,deck\,of\,cards}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{4}{52}=\frac{1}{13}\)

(ii) Total number cards in a deck = 52

Number of Queens in a deck of cards = 4

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,in\,a\,deck}\)

∴ Probability of drawing a queen from the deck of cards P(queen)

= \(\frac{Number\,of\,queen\,in\,a\,deck\,of\,cards}{Total\,number\,of\,in\,a\,deck}\) = \(\frac{4}{52}=\frac{1}{13}\)

(iii) Total number cards in a deck = 52 

Number of black cards in a deck of cards = 26 (13 spades and 13 clubs)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,in\,a\,deck}\)

∴ Probability of drawing a black card from the deck of cards P(black)

= \(\frac{Number\,of\,black\,cards\,in\,a\,deck\,of\,cards}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{26}{52}=\frac{1}{2}\)

(i) Total number cards in a deck = 52 

Number of kings in a deck of cards = 4

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of drawing a king from the deck of cards P(king)

= \(\frac{Number\,of\,kings\,in\,a\,deck\,of\,cards}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{4}{52}=\frac{1}{13}\)

(ii) Total number cards in a deck = 52 

Number of spades in a deck of cards = 13

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of drawing a spade from the deck of cards P(spade)

= \(\frac{Number\,of\,spades\,in\,a\,deck\,of\,cards}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{13}{52}=\frac{1}{4}\)

(iii) Total number cards in a deck = 52

Chances of drawing a Red queen from the deck of cards = 2 (they are queen of hearts and queen of diamonds)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of drawing a Red queen from the deck of cards P(Red queen)

= \(\frac{chances\,drawing\,a\,red\,queen\,from\,the\,deck\,of\,cards}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{2}{52}=\frac{1}{26}\)

(iv) Total number cards in a deck = 52 

Chances of drawing a black 8 from the deck of cards = 2 (they are 8 of clubs and 8 of spades)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of drawing a black 8 from a deck of cards P(black 8)

= \(\frac{chances\,drawing\,a\,black\,from\,from\,the\,deck\,of\,cards}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{2}{52}=\frac{1}{26}\)

(i) Total number of outcomes = 6 (they are 1,2,3,4,5,6) 

Chances of getting 2 on the die = 1

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting 2 on die P(2)

= \(\frac{possible\,chances\,of\,getting\,2}{Total\,number\,of\,outcomes}\) = \(\frac{1}{6}\)

(ii) Total number of outcomes = 6 (they are 1,2,3,4,5,6) 

Chances of getting a number less than 3 on the die = 2 (They are 1,2)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting a number less than 3 on die P(less than 3)

= \(\frac{Possible\,chances\,of\,getting\,a\,number\,less\,than\,3}{Total\,number\,of\,outcomes}\) = \(\frac{2}{6}=\frac{1}{3}\)

(iii) Total number of outcomes = 6 (they are 1,2,3,4,5,6) 

Composite number: A number which is not a prime number or a number which is divisible by numbers other than 1 and the number itself. 

Chances of getting a composite number on the die = 2 (They are 4,6)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting a composite number on die the P(composite number)

= \(\frac{Possible\,chances\,of\,getting\,a\,number\,less\,than\,3}{Total\,number\,of\,outcomes}\) = \(\frac{2}{6}=\frac{1}{3}\)

(iv) Total number of outcomes = 6 (they are 1,2,3,4,5,6) 

Chances of getting a number not less than 4 on the die = 4 (They are 4,5,6)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting a number not less than 4 on die P(not less than 4)

= \(\frac{Possible\,chances\,of\,getting\,a\,number\,less\,than\,3}{Total\,number\,of\,outcomes}\) = \(\frac{3}{6}=\frac{1}{2}\)

(i) Total number of ball bearings = 19 

Chances of drawing a prime numbered ball = 9 (They are 2,3,5,7,11,13,17,19)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of drawing a prime numbered ball bearing P(prime ball)

= \(\frac{possible\,chances\,of\,drawing\,a\,prime\,numbered\,ball\,bearing}{Total\,number\,of\,ball\,bearings}\) = \(\frac{8}{19}\)

(ii) Total number of ball bearings = 19 

Chances of drawing an even numbered ball = 9 (They are 2,4,6,8,10,12,14,16,18)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of drawing a prime numbered ball bearing P(prime ball)

\(\frac{possible\,chances\,of\,drawing\,a\,prime\,numbered\,ball\,bearing}{Total\,number\,of\,ball\,bearings}\) = \(\frac{9}{19}\)

(iii) Total number of ball bearings = 19 

Chances of drawing a numbered ball which is divisible by 3 = 6 (They are 3,6,9,12,15,18)

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of drawing a numbered ball bearing which is divisible by 3 P(ball divisible by 3)

= \(\frac{possible\,chances\,of\,drawing\,a\,numbered\,which\,is\,divisible\,by\,3}{Total\,number\,of\,ball\,bearings}\) = \(\frac{6}{19}\)

Here, U = {1, 2, 3, …, 120} ‘

A number is selected at random.

∴ Total number of primary outcomes is n = ¹²⁰C₁ = 120.

(1) A = Event that the number selected is a multiple of 3.

= {3, 6, 9, 12, …, 117, 120}

∴ Favourable outcomes for the event A is m = 40.

Hence, P(A) = \(\frac{m}{n} = \frac{40}{120} = \frac{1}{3}\)

(2) A’ = Event that the number selected is not a multiple of 3.

∴ P(A’) = 1 – P(A)

= \(1 – \frac{1}{3} = \frac{2}{3}\)

(3) B = Event that the number selected is a multiple of 4.

= {4, 8, 12, 16, …, 116, 120}

∴ Favourable outcomes for the event B is m = 30.

Hence, P(B) = \(\frac{m}{n} = \frac{30}{120} = \frac{1}{4}\)

(4) B’ = Event that the selected number is not a multiple of 4.

∴ P(B’) = 1 – P(B)

= 1 – \(\frac{1}{4} = \frac{3}{4}\)

(5) A ∩ B = Event that the number selected is a multiple of both 3 and 4, i.e., a multiple of 12.

∴ A ∩ B = {12, 24, 36, 48, …, 108, 120}

∴ Favourable outcomes for the event A ∩ B is m = 10.

Hence, P(A ∩ B) = \(\frac{m}{n} = \frac{10}{20} = \frac{1}{12}\)

To Find: Describe the given events. 

Explanation: when two dice are thrown then the no. of possible outcomes are 6² = 36 

Now, According to the question, 

A = Getting an even number of the first die. 

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

B = Getting an odd number on the first dice 

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 = Getting at most 5 as the sum of numbers on the two dices. 

C = {(1, 1)(1, 2)(1, 3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)} 

D = Getting a sum greater than 5 but less than 10 

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 = Getting at least 10 as the sum of numbers on the dices 

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

F = Getting an odd number on one of the dices 

{(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) Now, 

(i) A and B = AՌB = Փ 

Since, There is no common events in Both A and B so the intersection is Null (Փ). 

⇒ B or C = BUC 

BUC = {(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)(2, 1)(2, 2)(2, 3)(4, 1)} 

⇒ B and C = BՌC 

BՌC = {(1, 1), (1, 2)(1, 3)(1, 4)(3, 1)(3, 2)} 

⇒ A and E = AՌE 

AՌE = {(4, 6)(6, 4)(6, 5)(5, 6)(6, 6)} 

⇒ A or F = AUF 

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

(ii) 

(a) True, because AՌB = Փ 

(b) True, because AՌB = Փ and AUB = S 

(c) False, because AՌC = Փ 

(d) False, because CՌD = Փ but CUD≠S 

(e) True, because CՌDՌE = Փ and CUDUE = S 

(f) True, because A’ՌB’ = Փ 

(g) False, because AՌBՌF = Փ and AUBUF = S

Here, there are 6 letters R, A, N, D, O, M in the word RANDOM.

∴ The total number of ways of arranging these six letters is,

n = ₆P⁶ = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

A = Event that getting R in the first place and M in the last place.

∴ Favourable outcomes for the event A are obtained as follows:

∴ m = ₁P¹ × ₄P⁴ × ₁P¹

= 1! × 4! × 1!

= 1 × 24 ×1 = 24

Hence p(A) = \(\frac{m}{n} = \frac{24}{720} = \frac{1}{30}\)

Here, let B = child is male;

G = child is female.

The sample space of selecting one family from the families having

two children is expressed as follows:
U = {(B, B), (B, G), (G, B), (G, G)}

(1) Event A₁ = One child is a female

= {(B, G), (G, B)}

(2) Event A₂ = At least one child is a female

= {(B, G), (G, B), (G, G)}

Correct option is \((b) \frac{1}{n}\)

Correct option is (c) To measure the life of electric bulb

Two examples of random experiment are:

1. The experiment of throwing a balanced die and
2. The experiment of finding defective units from a lot of units produced.

Possible outcomes of a coin = HT

So here,

There is an equal possibility of coming of Head and Tail in every toss.

Now it is given that probability of getting a tail = 38

The probability of getting a head = 1 - probability of getting a tail \(=1-\frac{3}{8}=\frac{5}{8}\)

The coin is tossed 1000 times.

Hence number of times a head obtained \(=\frac{5}{8}\times 1000=625\) times

When two coins are tossed simultaneously then the possible outcomes obtained are {HH, HT, TH, and TT}.

Here H denotes head and T denotes tail.

Therefore, a total of 4 outcomes obtained on tossing two coins simultaneously.

The favourable outcome of getting at most one head are HT, TH, and TT.

So, the probability of getting at most one head is 3/4

(c) 0.42

P(Particular number comes on the dice) = \(\frac{1}{6}\)

(∵ there are in all 6 numbers)

P(particular number does not come on the dice) = 1 - \(\frac{1}{6}\) = \(\frac{5}{6}\)

As there are 3 dices, so,

P(Picked number does not come in any of dice) = \(\big(\frac{5}{6}\big)^3\)

P(You lose money) = \(\big(\frac{5}{6}\big)^3\) = \(\frac{125}{216}\) ≈ 0.58

∴ P(Winning) = 1 - 0.58 = 0.42

Total numbers of elementary events are: 7 

Let E be the event of having 53 Sundays 

Note- a non-leap year has 365 days or 52 weeks and 1 day. This one day could be any day amongst Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. Out of these favorable event is one; Sunday. 

Number of favorable outcome is = 1 

P (Sunday) = P (E) = \(\frac{1}7\)

Probability of the occurrence of an event cannot be greater than 1. 

1.6 Which is greater than 1 is the correct answer

Probability of an impossible event is always zero (0)

Correct option is: B) 1/2

Probability of a certain event is always 1.

C) Both (i) & (iv)

Correct option is (C) –0.2

\(\because\) For any event, \(0\leq P(A)\leq1\)

Hence, probability of any event never be negative.

Therefore, - 0.2 never be the probability of any event.

Correct option is  C) – 0.2

Total number of possible outcomes, n(S) = 12 {3 red balls, 5 black balls and 4 white balls) 

(i) n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)

(ii) n(E) = 3

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

(iii) n(E) = 5

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{12}\)

(iv) n(E) = 9

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{9}{12}\) = \(\frac{3}{4}\)

We know that 

Probability of occurrence of an event

= \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

By permutation and combination, total no. of ways to pick r objects from given n objects is ⁿC_r 

Now, total no. of ways to pick a ball from 20 balls is ²⁰c₁ = 20 

Our desired output is to pick a red ball. So, no. of ways to pick a red ball from 9 red balls (because the red ball can be picked from only red balls) is ⁹c₁ = 9

Therefore, the probability of picking a red ball = \(\frac{9}{20}\)

Conclusion: Probability of picking a red ball from 9 red, 

7 white and 4 black balls is  \(\frac{9}{20}\)

Class 9 Maths MCQ Questions of Probability with Answer are given here. Every one of the questions is arranged depending on the most recent exam pattern. Students can practice the part-savvy questions at Sarthaks eConnect and plan for the last, most important tests to score great marks. 

 Practicing Class 9 Maths important questions will help students to score good marks. After solving these MCQ Questions they get more confidence in the exams. 

Practice MCQ Questions for Class 9 Maths

1. When a die is thrown, the probability of getting an odd number less than 3 is 

(a) 1/6 
(b) 1/3 
(c) 1/2 
(d) 0

2. The probability of an event of a trial is always

(a) more than 1
(b) between 0 and 1 (both inclusive)
(c) 1
(d) 0

3. If two coins are tossed simultaneously, then what is the probability of getting exactly two tails?

(a) 1/4
(b) 1/2
(c) 1/3
(d) None of the above

4. Empirical probability is also known as 

(a) Classic probability
(b) Subjective probability
(c) Experimental probability
(d) None of the above

5. There are 4 green and 2 red balls in a basket. What is the probability of getting the red balls? 

(a) 1/2
(b) 1/3
(c) 1/5
(d) 1/6

6. Which of the following cannot be the probability of an event?

(a) 1
(b) 0
(c) 0.75
(d) 1.3

7. What is the probability of drawing a queen from the deck of 52 cards?

(a) 1/26
(b) 1/52
(c) 1/13
(d) 3/52

8. What is the probability that a leap year has 53 Sundays?

(a) 78
(b) 45
(c) 23
(d) 27

9. A card is selected from a deck of 52 cards. The probability of its being a red face card is 

(a) 3/26 
(b) 3/13 
(C) 2/13 
(D) 1/2

10.  An unbiased dice is thrown. What is the probability of getting an even number or a multiple of 3?
(a) 3/2
(b) 2/3
(c) 5/4
(d) 4/3

11. The sum of all probabilities equal to:

(a) 4
(b) 1
(c) 3
(d) 2

12. The probability of an impossible event is

(a) more than 1
(b) less than 1
(c) 1
(d) 0

13. Find the probability of a selected number is a multiple of 4 from the numbers 1, 2, 3, 4, 5, …15.

(a) 1/5
(b) 1/3
(c) 4/12
(d) 2/15

14. A card is drawn from a well-shuffled deck of 52 cards. What is the probability of getting a king of the red suits?

(a) 3/36
(b) 1/26
(c) 3/26
(d) 1/16

15. Performing an event once is called

(a) Sample
(b) Trial
(c) Error
(d) None of the above

16. What is the probability of getting an odd number less than 4, if a die is thrown?

(a) 1/6
(b) 1/2
(c) 1/3
(d) 0

17. Three coins were tossed 200 times. The number of times 2 heads came up is 72. Then the probability of 2 heads coming up is:

(a) 1/25
(b) 2/25
(c) 7/25
(d) 9/25

18. A batsman hits boundaries for 6 times out of 30 balls. Find the probability that he did not hit the boundaries.

(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5

19. If the probability of an event to happen is 0.3 and the probability of the event not happening is:

(a) 0.7
(b) 0.6
(c) 0.5
(d) None of the above

20. If P(E) = 0.38, then probability of event E, not occurring is:

(a) 0.62
(b) 0.38
(c) 0.48
(d) 1

Answer: 

1. Answer: (a) 1/6 

Explanation: When a die is thrown, then total number of outcomes = 6

Odd number less than 3 is 1 only.

Number of possible outcomes = 1

∴ Required probability = 1/6

2. Answer: (b) between 0 and 1 (both inclusive)

3. Answer: (a) 1/4

Explanation: If two coins are tossed, then the sample space, S = {HH, HT, TH, TT}

Favourable outcome (Getting exactly two tails) = {TT} 

Therefore, the probability of getting exactly two heads = 1/4

4. Answer: (c) Experimental probability

Explanation: Empirical probability is also known as experimental probability.

5. Answer: (b) 1/3

Explanation: Total balls = 4 green + 2 red = 6 balls

No. of red balls = 2.

Hence, the probability of getting the red balls = 2/6 = 1/3

6. Answer: (d) 1.3

Explanation: The probability of an event always lies between 0 and 1.

7. Answer: (c) 1/13

Explanation: Total cards = 52

Number of queens in a pack of 52 cards = 4

Hence, the probability of drawing a queen from a deck of 52 cards = 4/52 = 1/13.

8. Answer: (d) 27

Explanation: The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday}, Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.
So there are 7 possibilities out of which 2 have a Sunday. So the probability of 53 Sundays in a leap year is 2/7.

9. Answer: (a) 3/26 

Explanation: In a deck of 52 cards, there are 12 face cards i.e. 6 red (3 hearts and 3 diamonds) and 6 black cards (3 spade and 3 clubs) 

So, probability of getting a red face card = 6/52 = 3/26

10. Answer: (b) 2/3

Explanation: In a single throw of dice, we get 1,2,3,4,5 or 6.

No. of total outcomes =6

Even number or a multiple of 3 are 2,3,4,6

Favourable no. of outcomes = 4

P(even number or multiples of 3)= 4/6=2/3

11. Answer: (b) 1

12.  Answer: (d) 0

13. Answer: (a) 1/5

Explanation: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

Multiples of 4 from the sample space = {4, 8, 12}

Therefore, the probability of the selected number is a multiple of 5 is 3/15 = 1/5.

14. Answer: (b) 1/26

Explanation: In a pack of 52 cards, there are a total of 4 king cards, out of which 2 are red and 2 are black.

Therefore, in a red suit, there are 2 king cards. 

Hence, the probability of getting a king of red suits = 2/52 = 1/26.

15. Answer: (b) Trial

Explanation: Performing an event once is called a trial.

16. Answer: (c) 1/3

Explanation: Sample space, S = {1, 2, 3, 4, 5, 6}

Favourable outcomes = {1, 3}

Therefore, the probability of getting an odd number less than 4 = 2/6 = 1/3.

17. Answer: (d) 9/25

Explanation: Probability = 72/200 = 9/25

18. Answer: (d) 4/5

Explanation: No. of boundaries = 6

No. of balls = 30

No. of balls without boundaries = 30 – 6 =24

Probability of no boundary = 24/30 = 4/5

19. Answer: (a) 0.7

Explanation: Probability of an event not happening = 1 – P(E)

P(not E) = 1 – 0.3 = 0.7

20. Answer: (a) 0.62

Explanation: P(not E) = 1 – P(E) = 1-0.38 = 0.62

Click here to practice Probability MCQ Questions for Class 9 Maths

(ii) and (iv) can’t be the probability of occurrence of an event. So, (ii) and (iv) are the answers to our question. 

Explanation: We know that,

0 ≤ probability ≤ 1 i.e. probability can vary from 0 to 1(both are inclusive) So, (i) 0 can be possible as 0 ≤ probability ≤ 1

(ii)    \(-\frac{3}{4}\)  is not possible as it is less than 0 

(iii)    \(\frac{3}{4}\)  is possible as 0 ≤ probability  ≤ 1

(iv) \(\frac{4}{3}\) is not possible as it is greater than 1 

Conclusion:   \(\frac{3}{4}\)  and   \(-\frac{3}{4}\) are the probabilities that cannot occur.

(i) We know that,

Probability of occurrence of an event

   = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Total possible outcomes are alphabets from a to z 

Desired outcomes are a, e, i, o, u 

Total no. of outcomes are 26 and desired outputs are 5 

Therefore, the probability of picking a vowel = \(\frac{5}{26}\)

Conclusion: Probability of choosing a vowel is  \(\frac{5}{26}\)

(ii) We know that, 

Probability of occurrence of an event 

   = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Total possible outcomes are all alphabets from a to z

Desired outcomes are all alphabets except a, e, i, o, u 

Total no. of outcomes are 26 and desired outputs are 21 

Therefore, the probability of picking a consonant = \(\frac{21}{26}\)

Conclusion: Probability of choosing a consonant is  \(\frac{21}{26}\)