Class 10 Maths Chapter 14 Previous Year Questions - Probability

Ans: (a)
No. of red balls = 3 
No. of white balls = 5 
No. of black balls = 7 
Total balls = 15 
Probability that ball drawn is neither red nor black = 5/15 = 1/3

Ans: (c)
Probability of getting bad in the lot = 0.045 
Let the no. of bad eggs = x 
∴ Probability of bag eggs 

⇒ 0.045 = 400

⇒ 0.045 = x/400
⇒ x = 400 × 0.045
⇒ x = 18 
No. of bad eggs = 18 
No. of good eggs = 400 – 18 
= 382

Ans: (b)
Total outcomes, when two dice are thrown 
= {(1, 1)(1, 2)(1, 3)(1, 4)(1, 5)(1, 6)
(2, 1)(2, 2)(2, 3)(2, 4)(2, 5)(2, 6)
(3, 1)(3, 2)(3, 3)(3, 4)(3, 5)(3, 6)
(4, 1)(4, 2)(4, 3)(4, 4)(4, 5)(4, 6)
(5, 1)(5, 2)(5, 3)(5, 4)(5, 5)(5, 6)
(6, 1)(6, 2)(6, 3)(6, 4)(6, 5)(6, 6)}
Favourable outcomes= {(1, 2)(1, 3)...(2, 1), (2, 3)... (3, 1)(3, 2)(3, 4)...(4, 1)...
(6, 5)}
No. of favourable outcomes = 30
Total outcomes = 36
So, P(E) = 30/36
= 5/6

Ans: (b)
Total balls (outcomes) = 45 
No. of times boundaries hit = 9     
(E = hitting the boundry) = 9/45
= 1/5
∴ P(E= not hitting the boundary)

Ans: (a)
Total number of people = 20
Number of people who can't swim = 5
Number of people who can swim = 20 - 5 = 15
∴ Required probability = 15/20 = 3/4

Ans: (a)
Probability of happening of an event + Probability of non - happening of an event
∴ p +q = 1

Ans: (c)
Probability of winning first prize = Ticket bought by girl  / Total ticket sold
⇒ 0.08 = Ticket bought by girl  / 6000
⇒ Ticket bought by girl = 0.08 x 6000 = 480

Ans: (c)
Total number of outcomes = 6 x 6 = 36
Favourable outcomes are {1,4), (2, 5], (3, 6), (4, 1), |5. 2), (6.3) i.e.,  6 in number
∴ Required probability = 6/36 = 1/6

Ans: (d)
Total number of cards = 52
Number of ace card =4
∴ Number of non ace card = 52 - 4 = 48
∴ Required probability = 48/52 = 12/13

Ans: (c)
The leap year has 366 days, i.e, 52 weeks and 2 days.
∴ Required probability = 2/7
The non-leap year has 365 days. i.e.. 52 weeks and 1 day.
∴ Required probability = 1/7
Therefore, assertion is true but reason is false.

Ans: (b)
Probability of drawing a green ball = 3 x Probability of drawing a red ball

∴ n = 15 

Ans: Number of red balls =4
Number of blue balls = 3
Number of yellow balls = 2
Total number of balls = 4 + 3 + 2= 9
(i) P(drawing a red balI) = 4/9
(ii) P(drawing a yellow ball) = 2/9

Ans: Let A be the event of getting atmost one head, and 5 be the sample space.

S = (HH, HT, TH, TT) and A = (HT, TH, TT)

⇒ n(S) = 4
Also, n(A) = 3

Required probability = n(A) / n(S)
= 3/4

Ans: (b)
Sample space = {(H,H), (H,T), (T,H), (T,T)}
∴ Number of total outcomes = 4
Favourable outcomes = {(H,H)}
∴ Number of favourable outcomes = 1
∴ Required probability = 1/4

Ans: (a)
Sample space = (1, 2, 3, 4, 5, 6)
∴ Number of total outcomes = 6
Favourable outcomes = (4, 6)
∴ Number of favourable outcomes = 2
∴ Required probability = 2/6 = 1/3

Ans: (a)
We know that there are 52 complete weeks m 364 days
Since, it is non leap year.
So. there will be 52 Wednesdays and remaining 365^th day may be any of the days of week
So, total number of ways = 7

∴ Number of favourable outcomes = 1
∴ Required probability = 1/7

Ans: (c)
Total number of letters in the word MANGO are 5.
So, number of total outcomes = 5
Vowels in the word ‘MANGO’ are A, O
So, number of favourable outcomes = 2
∴ Required probability  = 2/5

Ans: (d)
Card numbered from {1,2, 3, ...., 20}
Total number of possible outcomes = 20
Odd prime numbers from 1 to 20 = {3, 5, 7, 11,13.17.19)
Total number of favourable outcomes = 7
Hence, the probability that the number on the card drawn is an odd prime number = 7/20

Ans: (d)
Total number of composite numbers between 1 to 20 = [4, 6, 8, 9, 10, 12, 14, 15, 16, 18)
Total number of favourable outcomes = 10
So, the probability that the number on the drawn card is a composite number = 10/20

∴ Required Probability = 1/2

Ans: (c)
Multiple of 3 = {3, 6, 9,12,15, 18}
Multiple of 6 = (6, 12, 18)
Multiple of 9 = (9, 18)
Total number of favourable outcomes = 1
Hence the probability that the card is a multiple of 3, 6 and 9 = 1/20
∴ Required Probability = 1/20

Ans: (c)
 Multiple of 3 between 1 to 20 = {3, 6, 9, 12,15, 18}
Multiple of 7 between 1 to 20 = (7,14)
∴ Multiple of 3 and 7 = 0
∴ Total number of favourable outcomes = 0
∴ Required Probability = 0

Ans: (b)
If all odd number cards are removed then remaining cards which are left = {2,4, 6, 8,10,12,14,16, 18, 20}
Now, prime number cards in remaining cards = 1
So, the probability of getting a prime number from the remaining cards = 1/10

Ans: The probability of an event that is sure to happen is 1.

Ans: Given, P(E) =0.023
 = 1- P(E) = 1 - 0.023 = 0.977

Ans: Total number of English alphabets = 26
Number of consonants = 26 - 5 = 21
∴ Number of favourable outcomes = 21
P (chosen letter is a consonant) = 21/26

Ans: Total number of outcomes = 6
Favourable outcomes are {1.2} i.e.. 2 in number
∴ Required probability = 2/6 = 1/3

Ans: Given, probability of winning a game is 0.07
∴  Probability of losing it = 1 - 0.07 = 0.93

Ans: There a re 18 marbles in the jar.
∴ Number of possible outcomes = 18
Let there are x yellow marbles in the jar.
∴ Number of red marbles = 18 - x
⇒ Number of favourable outcomes = (18 - x)
∴ Probability of drawing a red marble = (18 - x) / 18
Now. according to the question, = (18 - x) / 18 = 2/3
⇒ 3(18 - x ) = 2 x 13
⇒ 54 -3x  = 36
⇒ 3x = 18
⇒ x  = 6
So, number of ye How marbles in jar = 6

Ans: Since, throwing a die twice or throwing two dice simultaneously are same.
Possible outcomes are:

(i) Let N be the event t hat 5 wiII come up at least once, the n number of favourable outcomes 
= 5 + 6 
= 11

(ii) Let E be the event that 5 does not come up either time, then number of favourable outcomes 
= [36 - (5 + 6)]
= 25

Ans: Total number of outcomes = {-3, -2, -1,0, 1, 2, 3} i.e. 7.
∴ Number of favourable outcomes = (4, 1, 0, 1, 4) i.e., 5.
∴ Required Probability = 5/7

Ans: Total number of possible outcomes = 36 
Only one outcome, i.e., (1, 1) has the product of the two numbers as 1. 
So, the required probability is 1/ 36 .

Ans: (A) Of the 600 plants, there are 150 fruit plants and 110 flowering plants.
So, required probability
=
(B) Of the 600 plants, there are 290 (125 + 165) plants which are either neem plants or peepal plants.
So, required probability = 290/600 i.e., 29/60

Ans: All possible outcomes are –3, –2, –1, 0, 1, 2, 3 
Favourable outcomes are – 1, 0, 1 (As x² < 4) 
So, required probability = 3/7

Ans: A leap year has 52 complete weeks + 2 days. 
These two days may be 
(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat) and (Sat, Sun). 
Of the 7 possible outcomes, only 1 
i.e., (Sun, Mon) is the favourable outcome. 
So, required probability is 1/7

Ans: (A) Shweta will be allowed to pick up a marble, only when the spinner stops on an even number. 
P(getting an even number) = 5 / 6 
Hence, the probability that she will be allowed to pick a marble from the bag is 5 / 6 
(B) P (getting a black marble) = 6 / 20 , or 3 / 10 . 
∴ Probability of getting a prize is 3 / 10 .

Ans: Cards are numbered from 7 to 40. i.e. {7,8,9, ......, 40}
So, total number of outcomes = 34
Multiple of 7 lies between 7 to 40 are {7, 14, 21, 28, 35}
∴ Total number of favourable outcomes= 5
∴ Required probability = 5/34

Ans: Total number of cards = 52
Total number of spade cards = 13
Total number of king cards = 4
Total number of spade cards and king cards = 13 + 4 - 1 = 16
[One card is subtracted as it is already included as a king of spade]
∴ Probability of drawing a spade or king card = 16/52
So, probability of drawing a card which is neither a spade nor a king = 1- 16/52
= 9/13

Ans: If a pair of dice is thrown once, then possible outcomes are:

∴ Number of possible outcomes are 36.
(i) Total possible outcomes of getting even number on each die
= {( 2, 2), ( 2, 4 ), ( 2, 6 ), ( 4, 4 ), [4, 6), (6, 6). (6, 2), (6, 4 ), (4, 2)}
Number of favourable outcomes = 9
∴ Required probability of getting an even number on each die = 9/36 = 1/4
(ii) Total possible outcomes of getting a total of 9
= {(3, 6), (4, 5), ( 5, 4), (6, 3)} which are 4 in number.
∴ Probability of getting a total of 9 = 9/36 = 1/4

Ans: We have, total number of balls = x + 2x + 3x = 6x
Total number of outcomes =6x
 (i) Number of favourable outcomes = 3x
∴ Probability of getting red ball = 3x /6x = 1/2
Now, probability of not getting red ball = 1-1/2 = 1/2
∴ Requited probability = 1/2
(ii) Total nu m be r of favourable outcomes = x
∴ Probability of getting white ball = x / 6x
∴ Required probability = 1/6

Ans: Total possible outcomes are f 1, 2, 3, 4, 5, i.e., 6 in number.
(i) Favourable outcomes are {2, 3, 5} i.e.. 3 in number.
∴ P (getting a prime number) = 3/6 = 1/2
(ii) Favourable outcomes are {3, 4, 5} i.e., 3 in number.
∴ P(getting a number lying between 2 and 6) = 3/6 = 1/2

Ans: When a coin is tossed 3 times, then total possible outcomes are 
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ Total number of possible outcomes = 8
Possible outcomes to lose the game are {HHT, HTH, THH, HTT,THT, TTH}
∴ Number of favourable outcomes = 6
∴ Required Probability = 6/8 = 3/4

Ans: Total number of cards = 50 - 5 + 1 = 46 
∴ Total number of possible outcomes = 46
(i) Prime numbers less than 10 are 5, 7.
So, number of favourable outcomes = 2
∴ P(getting a prime number less than 10) = 2/46 = 1/23
(ii) Per feet squares from 5 to 50 are 9, 16, 25,  36, 49 i.e., 5 in number.
∴ P (getting a number which is a perfect square) =5/46

Ans: Total number of faces in a die = 6
(i) Number of favourable outcomes = 3
∴ P(getting A) = 3/6 = 1/2
(ii) Number of favourable outcomes = 2
∴ P (getting B)  = 2/6 = 1/3

Ans: Total number of outcomes = 20 
Let, E be the event that a number selected is a prime number. 
Since, the prime number between 1 to 20 (or favourable cases) are 2, 3, 5, 7, 11, 13, 17, 19 
∴ Number of favourable outcomes = 8
∴ P(E) = Number of  favourable outcomes /  Total number of outcomes
= 8/20 = 2/5
Hence, the required probability is 2/5

Ans: The given numbers are {–3, –2, –1, 0, 1, 2, 3} 
The square of these numbere are {9, 4, 1, 0, 1, 4, 9} 
∴ Total numbers of outcomes = 7 
The square of numbers that are less than or equal to 1 = {1, 0, 1}
∴ Number of favourable outcomes = 3 
P(getting a square of a number less than or equal to 1) = 3 / 7 
Hence, the required probability is 3 / 7 .

Ans: (A) Total number of ball pens = 144 
∴Total number of outcomes is 144. 
Also, the number of defective ball pens = 20 
∴ Non-defective ball pens = 144 – 20 = 124 (A) 
Let E1 be the event that customer will buy a ball pen i.e., ball pen is non-defective. 
∵Total number of non-defective pens = 124

Hence, the probability that customer will buy the ball pen is 31 / 36 .
(B) Probability of not buying the ball pen 
= 1 – Probability of buying the ball pen 
= 1 – P(E₁) 
= 1 – 31 / 36 
= 5 / 36 
Hence, the probability that the customer will not buy the ball pen is 5 / 36

Ans: Number of cards removed = (2 + 2 + 2 + 2) = 8 
Total number of remaining cards = (52 – 8) = 44 
Now, there are 2 jacks and 2 kings of black colour and 2 queens and 2 aces of red colour left. 
(A) Number of black queens = 0 
∴ P(getting a black queen) = 0 / 44 = 0 
(B) Number of red cards = 26 – 4 = 22 
∴ P(getting a red card) = 22 / 44 = 1/ 2 
(C) Number of jacks of black colour = 2 
∴ P(getting a black jack) = 2 / 44 = 1 / 22
(D) We know that jacks, queens and kings are face cards. 
∴ Number of remaining face cards = (2 + 2 + 2) = 6 
∴ P(getting a face card) = 6/44 = 3 / 22

Ans:  For a / b > 1, 
When a = 1, b can not take any value. 
When a = 2, b can take one value i.e., 1.
When a = 3, b can take two values i.e., 1, 2. 
When a = 4, b can take three values i.e., 1, 2, 3. 
When a = 5, b can take four values i.e., 1, 2, 3, 4. 
When a = 6, b can take five values i.e., 1, 2, 3, 4, 5. 
Here, total number of possible outcomes is same as when we throw a dice twice. 
∴ Total possible outcomes = 36

∴
= 15/36
= 5 / 12
Hence, the required probability is 5 / 12