Worksheet Solutions: Probability

# Probability Class 10 Worksheet Maths Chapter 14

Multiple Choice Questions
Q1: The probability of an impossible event is
(a) 0.01
(b) 100
(c) zero
(d) 1

Ans: (a)
Number of kings = 3 (one spade king is counted in No. of spades)
Number of possible outcomes = 13 + 3 = 16
Number of Total outcomes = 52
Required Probability =16/52 = 4/13

Q2: The probability that a leap year will have 53 Sundays or 53 Mondays is
(a) 4/7
(b) 2/7
(c) 1/7
(c) 3/7

Ans: (d)
Leap year contains 366 days = 52 weeks + 2 days 52 weeks contain 52 Sundays and 52 weeks contain 52 Mondays
We will get 53 Sundays or 53 Mondays if one of the remaining two days is a Sunday or Monday Total possibilities for two days are:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday,Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
Number of Total possible outcomes = 7
Number of possible outcomes either Sunday or Monday or Both = 3
Required Probability = 3/7

Q3: A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and these values are equally likely outcomes. The probability that it will point at a number greater than 5 is
(a) 1/2
(b) 1/4
(c) 1/5
(d) 1/3
Ans:
(a)
Number of possible outcomes = {6, 7, 8, 9, 10} = 5
Number of total outcomes = 10
∴ Required Probability = 5/10 =1/2

Q4: The probability of an impossible event is
(a) 0.01
(b) 100
(c) zero
(d) 1
Ans:
c
An event which has no chance of occurrence is called an impossible event.
for example: The probability of getting more than 6 when a die is thrown is an impossible event because the highest number in a die is 6
The probability of an impossible event is always zero.(0)

Q5: Cards marked with numbers 1, 2, 3, ______, 25 are placed in a box and mixed thoroughly and one card is drawn at random from the box. The probability that the number on the card is a multiple of 3 and 5 is
(a) 12/25
(b) 4/25
(c) 1/25
(d) 8/25

Ans: (c)
Multiples of 3 = 3 6 9 12 15 18 21 24
Multiples of 5 = 5 10 15 20 25
Number of possible outcomes (multiple of 3 and 5) = {15} = 1
Number of Total outcomes = 25
∴ Required Probability =1/25

Q6: A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and these values are equally likely outcomes. The probability that it will point at a number greater than 5 is
(a) 1/2
(b) 1/4
(c) 1/5
(d) 1/3

Ans: a
Number of possible outcomes = {6, 7, 8, 9, 10} = 5
Number of total outcomes = 10
∴ Required Probability = 5/10 =1/2

Q7: A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. Find the probability that it bears a two digit number
Ans:
81/89

Q8: A card is drawn from a pack of 52 cards. Find the probability of getting a king of red colour

Ans: 1/26

Q9: A bag contains 40 balls out of which some are red, some are blue and remaining are black. If the probability of drawing a red ball is 11/20 and that of blue ball is 1/5 , then what is the no. of black balls?
Ans:
10

Q10: A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the drawn card is neither a king nor a queen.
Ans:
Total number of cards = 52.
Total number of kings and queens = 4+4 = 8.
Remaining number of cards = 52 - 8 = 44.
P(getting a card which is neither a king nor a queen) =44/52 =11/13

Q11: Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.
Ans:
We know that ,When a pair of dice is thrown, the total number of possible outcomes are 36.
Favourable outcomes of the sum greater than 10 = {(5,6)(6,5)(6,6)}
Therefore, number of cases favourable to the event = 3
Hence, Probability of getting the sum greater than 10 = 3/36 = 1/12

Q12: Three unbiased coins are tossed together. Find the probability of getting at least two heads?
Ans:
If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that the event "Getting at least two heads" occurs.
Favourable number of elementary events = 4
total no. of possible events when three coins are tossed = 8
Hence, required probability = 4/8 = 1/2

Q13: Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is
(1) a prime number less than 10

(2) a number which is a perfect square.
Ans:
According to question we are given that Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random.
Therefore All possible outcomes are 5, 6, 7, 8 …………… 50.
Number of all possible outcomes = 46
(1) Out of the given numbers, the prime numbers less than 10 are 5 and 7.
Suppose E1 be the event of getting a prime number less than 10.
Then, number of favorable outcomes = 2
Therefore, P(getting a prime number less than 10) = P(E) = 2/46 = 1/23
(2) Out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.
Suppose E2 be the event of getting a perfect square.
Then, number of favorable outcomes = 5
Therefore, P(getting a perfect square) = P(E) = 5/46

Q14: Why is tossing a coin considered as the way of deciding which team should get the ball at the beginning of a football match?
Ans:
Proabibilty of the event  = Number of favourable outcomes / Total Number of possible outcomes
Probability of head= P(H) = 1/2
Probability of tail = P(T) = 1/2
i.e. P (H) = 1/2 = P (T) = 1/2
Probability of getting head and tail both are same.
Tossing a coin considered to be fairway.

Q15: Two coins are tossed together. Find the probability of getting both heads or both tails.
Ans:
Two coins are tossed together.
Possibilities are HH, HT, TH, TT
Total outcomes = 4
Both heads or both tails = HH, TT
Number of favourable outcome = 2
Probability = Number of favourable outcomes / Total Number of outcomes
P(HH or TT) = 2/4 = 1/2

Q16: A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is 3/10 and that of a black ball is , 2/5 then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.
Ans:
P (White ball) = 3/10
P(Black ball) = 2/5
P(E) = 1 - P( not E)
Probability of not getting white and black ball is equals to probability of getting red balls.
P(Red ball) = 1 - (3/10 + 2/5) = 3/10
2/5 x Total no of balls = 20 (red balls)
Hence, Total numbers of balls = 20x5/2 = 50

Q17: A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(1) a two digit number,
(2) number divisible by 5.

Ans: 20. No. of all possible outcomes = 90
(1) Discs with two digit number are 10 to 90
No. of discs with two digits numbers = 90 - 10 +1 = 81
∴ No of favourable outcomes = 81
P(a disc with two digit number) = No of favourable outcome / No of all possible outcome = 81/90 = 9/10
(2) The numbers having 0 or 5 at its once place are divisible by 5 = 5,10,15.....90
Total no. of favourable outcomes = 18
P(a disc with a number divisible by 5) = 18/90 = 1/5

Q18: A box contains cards bearing numbers 6 to 70. If one cards is drawn at random from the box, find the probability that it bears
(1) a one digit number.
(2) a number divisible by 5,
(3) an odd number less than 30,

Ans: (1) Let E be the event of getting a one digit number.
Number of possible outcomes = 70 - 6 + 1 = 65
The outcomes favourable to E are 6,7,8 and 9
Number of favourable outcomes = 4
P(E)=P(Getting a one digit number) = 4/65
(2) Let F be the event of getting a number divisible by 5.
Number of possible outcomes =65
The outcomes favourable to the event F are 10,15,20,...., 65,70.
Number of outcomes favourable to F = 13
P(F)=P(Getting a number divisible by 5) = 13/65 = 1/5
(3) Let G be the event of getting an odd number less than 30.
Number of possible outcomes =65
The outcomes favourable to the event G are 7,9,11,13,....., 29.
Number of favourable outcome =12
P(G)=P(Getting an odd number less than 30) = 12/65

Q19: All red face cards are removed from a pack of playing cards. The remaining cards are well-shuffled and then a card is drawn at random from them. Find the probability that the drawn card is
(1) a red card,
(2) a face card,
(3) a card of clubs.
Ans:
There are 6 red face cards. These are removed.
Thus, remaining number of card = 52 – 6 = 46.
(1) Number of red cards now = 26 – 6 = 20.
Therefore, P(getting a red card) = Number of favourable outcomes / Number of all possible outcomes = 20/46 = 10/23
Thus, the probability that the drawn card is a red card is 10/23.
(2) Number of face cards now = 12 – 6 = 6.
Therefore, P(getting a face card) = Number of favourable outcomes / Number of all possible outcomes = 6/46 = 3/23
Thus, the probability that the drawn card is a face card is 3/23.
(3) The number of card of clubs = 12.
Therefore, P(getting a card of clubs) = Number of favourable outcomes / Number of all possible outcomes = 12/46 = 6/23
Thus, the probability that the drawn card is a card of clubs is 6/23

Q20: Cards numbered 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is (i) an odd number, (ii) a perfect square number, (iii) divisible by 5, (iv) a prime number less than 20.
Ans:
According to the question,
All possible outcomes are 11. 12. 13,....... 60
Total number of possible outcomes = (60 - 10) = 50
i. Suppose, E1 be the event that the number on the drawn card is an odd number.
⇒ the favourable outcomes are 11. 13. 15,....... 59
Clearly, these numbers form an AP with a = 11 and d = 2.
Tn = 59 ⇒ 11 + (n - 1) x 2 = 59 ⇒ (n - 1) x 2 = 48 ⇒ n - 1 = 24 ⇒ n = 25
So, the number of favourable outcomes = 25
∴ P(getting an odd number) = P (E1) = 25/50 = 1/2 .
ii. Suppose, E2 be the event that the number on the drawn card is a perfect square number.
⇒ the favourable outcomes are . 16. 25. 36. 49..
The number of favourable outcomes = 4.
∴ P(getting a perfect square number) P (E2) = 4/50 = 2/25 ..
iii. Suppose, E3 be the event that the number on the drawn card is divisible by 5.
⇒ the favourable outcomes are 15, 20. 25..........60
Clearly, these numbers form an AP with a = 15 and d=5.
Tm = 60 ⇒ 15 + (m - 1) x 5 = 60 ⇒ (m - 1) x 5 = 45 ⇒ m - 1 = 19 ⇒ m = 10
So, the number of favourable outcomes = 10
∴ P(getting a number divisible by 5) = P (E3) = 10/50 = 1/5
iv. Suppose, E4 be the event that the number on the drawn card is a prime number less than 20.
⇒ the favourable outcomes are 11, 13, 17, 19
So, the number of favourable outcomes = 4
∴ P(getting a prime number less than 20) = P (E4) = 4/50 = 2/25

The document Probability Class 10 Worksheet Maths Chapter 14 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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