Two dice, one blue and one grey, are thrown at the same time. Complete the following table: Event: ‘Sum on two dice' 2 3 4 5 6 7 8 9 10 11 12 Probability From the above table a student argues that there are 11 possible outcomes 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 and 12 . Therefore, each of them has a probability 1 11 . Do you agree with this argument?

It is given that one blue and one grey dice are thrown at the same time. Let us first write all the possible outcomes that can occur, S = ( 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 ) Total number of possible outcomes when two dice are thrown, n ( S ) = 36 Sum of the numbers on pair of dice is 2 : Favourable cases when sum of the numbers on pair of dice is 2 , E = ( 1 , 1 ) Number of favourable outcomes, n ( E ) = 1 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 1 36 Hence, the probability of getting 2 as the sum on pair of dice is 1 36 . Sum of the numbers on pair of dice is 3 : Favourable cases when sum of the numbers on pair of dice is 3 , E = ( 1 , 2 ) , ( 2 , 1 ) Number of favourable outcomes, n ( E ) = 2 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 2 36 = 1 18 Hence, the probability of getting 3 as the sum on pair of dice is 1 18 . Sum of the numbers on pair of dice is 4 : Favourable cases when sum of the numbers on pair of dice is 4 , E = ( 1 , 3 ) , ( 3 , 1 ) , ( 2 , 2 ) Number of favourable outcomes, n ( E ) = 3 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 3 36 = 1 12 Hence, the probability of getting 4 as the sum on pair of dice is 1 12 . Sum of the numbers on pair of dice is 5 : Favourable cases when sum of the numbers on pair of dice is 5 , E = ( 2 , 3 ) , ( 3 , 2 ) , ( 1 , 4 ) , ( 4 , 1 ) Number of favourable outcomes, n ( E ) = 4 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 4 36 = 1 9 Hence, the probability of getting 5 as the sum on pair of dice is 1 9 . Sum of the numbers on pair of dice is 6 : Favourable cases when sum of the numbers on pair of dice is 6 , E = ( 2 , 4 ) , ( 4 , 2 ) , ( 3 , 3 ) , ( 1 , 5 ) , ( 5 , 1 ) Number of favourable outcomes, n ( E ) = 5 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 5 36 Hence, the probability of getting 6 as the sum on pair of dice is 5 36 . Sum of the numbers on pair of dice is 7 : Favourable cases when sum of the numbers on pair of dice is 7 , E = ( 2 , 5 ) , ( 5 , 2 ) , ( 3 , 4 ) , ( 4 , 3 ) , ( 1 , 6 ) , ( 6 , 1 ) Number of favourable outcomes, n ( E ) = 6 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 6 36 = 1 6 Hence, the probability of getting 7 as the sum on pair of dice is 1 6 . Sum of the numbers on pair of dice is 8 : Favourable cases when sum of the numbers on pair of dice is 8 , E = ( 4 , 4 ) , ( 3 , 5 ) , ( 5 , 3 ) , ( 2 , 6 ) , ( 6 , 2 ) Number of favourable outcomes, n ( E ) = 5 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 5 36 Hence, the probability of getting 8 as the sum on pair of dice is 5 36 Sum of the numbers on pair of dice is 9 : Favourable cases when sum of the numbers on pair of dice is 9 , E = ( 6 , 3 ) , ( 3 , 6 ) , ( 5 , 4 ) , ( 4 , 5 ) Number of favourable outcomes, n ( E ) = 4 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 4 36 = 1 9 Hence, the probability of getting 9 as the sum on pair of dice is 1 9 Sum of the numbers on pair of dice is 10 : Favourable cases when sum of the numbers on pair of dice is 10 , E = { ( 4 , 6 ) , ( 5 , 5 ) , ( 6 , 4 ) } Number of favourable outcomes, n ( E ) = 3 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 3 36 = 1 12 Hence, the probability of getting 10 as the sum on pair of dice is 1 12 . Sum of the numbers on pair of dice is 11 : Favourable cases when sum of the numbers on pair of dice is 11 , E = ( 6 , 5 ) , ( 5 , 6 ) Number of favourable outcomes, n ( E ) = 2 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 2 36 = 1 18 Hence, the probability of getting 11 as the sum on pair of dice is 1 18 . Sum of the numbers on pair of dice is 12 : Favourable cases when sum of the numbers on pair of dice is 12 , E = ( 6 , 6 ) Number of favourable outcomes, n ( E ) = 1 We know that, Probability , P ( E ) = n ( E ) n ( S ) = 1 36 Hence, the probability of getting 12 as the sum on pair of dice is 1 36 . Event: 'Sum on two dice' 2 3 4 5 6 7 8 9 10 11 12 Probability 1 36 1 18 1 12 1 9 5 36 1 6 5 36 1 9 1 12 1 18 1 36 Since, the probability of the sum of pair of dice for given numbers is not same. Hence, each of them does not have probability of 1 11 .

MEDIUM

10th CBSE

IMPORTANT

Earn 100

Rehan has a set of $40$ cards numbered $1 - 40$ without repeating any number. Rehan wants to construct $△ PQR$ shown below by choosing the length of $PR$ , in $cm$ , from the set of the cards.

What is the probability that a card chosen at random makes $△ PQR$ an OBTUSE angled triangle with $Q$ as obtuse angle?

Important Questions on Probability

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 2

Five different letters are put inside

5

addressed envelopes by an illiterate servant. What is the probability that only

2

letters are placed in the correct envelopes?

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 3

A friend is going to roll a six sided die and says “if I roll and even number we will play a checkers. if I roll an odd number we will play chess”. She States “I know we are going to play chess because it is more likely I will roll an odd number”.Is your friend correct? Which is more likely rolling an odd number or an even number? show your probability calculation as fraction and percentage. use your calculation to confirm or refuse your friends claim.

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 4

Two dice are numbered

1, 2, 3, 4, 5, 6

and

1, 1, 2, 2, 3, 3

respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from

2

9

separately.

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 5

A bag contains

12

blue balls and

x

red balls. If one ball is drawn at random. What is the probability that it will be a red ball? If

8

more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in the first case, then find

x

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 6

Two dice are numbered $1, 2, 3, 4, 5, 6$ and $1, 1, 2, 2, 3, 3$ respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from $2$ to $9$ , separately.

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 7

Two dice, one blue and one grey, are thrown at the same time. Complete the following table:

Event: ‘Sum on two dice'	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability

From the above table a student argues that there are

11

possible outcomes

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

and

12

. Therefore, each of them has a probability

\frac{1}{11}

. Do you agree with this argument?

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 8

Two customers are visiting a particular shop in the same week (Monday to Saturday). Each person is equally likely to visit the shop on any one day as the other. What is the probability that both will visit the shop on different days?

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 15 - Probability>Exercise>Q 9

A bag contains

5

red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Rehan has a set of 40 cards numbered 1-40 without repeating any number. Rehan wants to construct △PQR shown below by choosing the length of PR, in cm, from the set of the cards. What is the probability that a card chosen at random makes △PQR an OBTUSE angled triangle with Q as obtuse angle?

Important Questions on Probability

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