Probability Class 10 Important Questions
The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.
Question 4.
In a single throw of a pair of different dice, what is the probability of getting
- a prime number on each dice?
- a total of 9 or 11?
Solution:
Total possible cases when two dice are thrown together = 6 x 6 = 36
Favourable cases when both numbers are prime are (2, 2), (2, 3), (2, 5), (3, 2), (3, 3),(3, 5), (5, 2), (5, 3), (5,5), i.e. 9 outcomes
![Probability Class 10 Maths CBSE Important Questions With Solutions 4](https://i1.wp.com/www.cbsetuts.com/wp-content/uploads/2020/07/Probability-Class-10-Maths-CBSE-Important-Questions-With-Solutions-4.png?resize=463%2C46&ssl=1)
(ii) Favourable cases when sum of numbers are 9 or 11 are (3, 6), (4, 5), (5, 4), (5, 6),(6, 3), (6,5), i.e. 6 outcomes
![Probability Class 10 Maths CBSE Important Questions With Solutions 5](https://i2.wp.com/www.cbsetuts.com/wp-content/uploads/2020/07/Probability-Class-10-Maths-CBSE-Important-Questions-With-Solutions-5.png?resize=422%2C44&ssl=1)
Question 5.
Two different dice are thrown together. Find the probability of:
- getting a number greater than 3 on each die.
- getting a total of 6 or 7 of the numbers on two dice
Solution:
- When two dice are thrown together total possible outcomes = 6 X 6 = 36
Favourable outcomes when both dice have number more than 3 are (4, 4), (4, 5),(4, 6), (5, 4), (5, 5), (5, 6), (6,4), (6,5), (6, 6), i.e. 9 outcomes.
=9/36=1/4 - Favourable outcomes when sum of the numbers appearing on the dice is 6 or 7 are, i.e. (1,5), (1, 6), (2,4), (2,5), (3,3), (3,4), (4, 2), (4, 3), (5,1), (5,2), (6,1), outcomes.
P(a total of 6 or 7) =11/36
Question 6.
A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal’ another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that
- Ramesh will buy the selected shirt?
- ‘Kewal’ will buy the selected shirt?
Solution:
- When one shirt is taken out, then number of total possible outcomes = 100
Ramesh will purchase when shirt is good,
Favourable outcomes = number of good shirts = 88 - Kewal will buy shirt if a shirt is not having major defect.
Number of favourable outcomes = Number of shirts without major defect = 96
P(Kewal buys a shirt) =96/100=24/25
Question 7.
Three different coins are tossed together. Find the probability of getting
- exactly two heads
- at least two heads
- at least two tails.
Solution:
Possible outcomes when three coins are tossed HHH, HHT, HTT, TTT, THH, TTH, HTH, THT
- Number of exactly two heads are HHT, HTH and THH.
P(exactly two heads) = 3/8 - In case of at least two heads, outcomes are HHT, HTH, THH and HHH.
P(at least two heads) = 4/8=1/2 - In case of at least two tails, outcomes are TTH, THT, HTT and TTT.
P(at least two tails) = 4/8=1/2
Question 8.
From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:
- a black king.
- a card of red colour.
- a card of black colour.
Solution:
Removed red colour cards = 3×2 = 6
Remaining cards = 52 – 6 = 46
- Number of black kings = 2
P(a black king) = 2/46=1/23 - Number of red colour cards = 26
Remaining red colour cards = 26 – 6 = 20
P(a card of red colour) = 20/46=10/23 - Number of black cards = 26
P(a black colour card) =26/46=13/23
Question 9.
There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card:
- is divisible by 9 and is a perfect square.
- is a prime number greater than 80.
Solution:
Total possible cases = 100
- Favourable cases when number is a perfect square and is divisible by 9 are 9, 36 and 81.
So, number of favourable cases = 3 - Favourable cases the prime numbers greater than 80 are 83, 89 and 97
So, number of favourable cases = 3
Question 10.
A game consist of tossing a one-rupee coin 3 times and nothing the outcome each time. Ramesh will win the game if all the show the tosses same result, (i.e. either all three heads or all three tails) and loses the game otherwise. Find the probability that Ramesh will lose the game.
Solution:
When three coins are tossed together, then total outcomes are HHH, HHT, HTT, TTT, TTH, THH, HTH, THT Total possible cases = 8
Favourable cases to win the game are HHH or TTT, i.e. two cases.
Required probability = P(Ramesh will loose the game) = 1 -1/4=3/4
Long Answer Type Questions [4 Marks]
Question 11.
A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1,2,3,…, 8 which are equally likely outcomes. What is the probability that the arrow will point at
- an odd number
- a number greater than 3
- a number less than 9.
Solution:
- Total possible outcomes when the arrow points at one of the numbers are 8.
Favourable outcomes when the required number is odd are 1, 3, 5, 7, i.e. 4 outcomes. - Favourable outcomes when the required number is more than 3 are 4,5,6,7, 8, i.e. 5 outcomes.
- Favourable outcomes when the required number is less than 9 are 1,2,3,4,5,6,7,8 i.e 8 outcomes
Question 12.
A number JC is selected at random from the numbers 1,2,3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of JC and y is less than 16.
Solution:
x can be any one of 1, 2, 3 or 4 andy can be any one of 1, 4, 9 or 16.
Total number of cases of xy = 16
Number of cases when product is less than 16 are 1 x 1,1 x 4,1 x 9, 2 x 1, 2 x 4, 3 x 1, 3 x 4,4 x 1, i.e. 8 cases.
Question 13.
A number x is selected at random from the numbers 1,4,9,16 and another numbery is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16
Solution:
x can be 1, 4, 9 or 16 andy can be 1, 2,3 or 4.
Total number of cases of Ay are 16.
Number of cases when Ay is more than 16 are (9 x 2), (9 x 3), (9 x 4), (16 x 2), (16 x 3), (16 x 4), i.e. 6 cases.
6 3
P(value of xy more than 16) =6/16=3/8
Question 14.
In figure is shown a disc on which a player spins an arrow twice. The fraction a/b is formed, where ‘a’ is the number of sector on which arrow stops on the first and b is the numberof the sector in which the arrow stops on second spin. On each spin, each sector has equal chance of selection by the arrow. Find the probability that the a/b> 1.
Solution:
For a/b > 1, when a = 1, b can not take any value
a = 2, b can take 1 value
a = 3, b can take 2 value, i.e. 1 and 2
a = 4, b can take 3 value, i.e. 1, 2, 3
a = 5, b can take 4 value, i.e. 1, 2, 3,4
a = 6, b can take 5 value, i.e. 1,2,3,4,5
Total possible outcomes = 6 x 6 = 36
Favourable outcomes = 1 + 2 + 3 + 4 + 5 = 15