Answer
Verified
498.9k+ views
HINT- In order to solve such type of question we must use formula Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] , along with proper understanding of favourable cases and total cases.
Complete step-by-step answer:
Here we have one unbiased dice which is numbered from 1 to 6. While other dice are biased on which numbers are 1,1,2,2,3,3.
When we through both the dice then every number of each dice will make a pair with each number of other dice so total numbers of possible outcomes will be
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
So total Favourable outcomes =36
The probability of getting each sum from 2 to 9
(1) Sum of 2
favourable outcome for Sum of 2 will be (1,1,), (1,1).
No. of favourable outcomes = 2
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 2)=$\dfrac{2}{{36}}{\text{ = }}\dfrac{1}{{18}}$
(2) Sum of 3
a favourable outcome for Sum of 3 will be (1,2), (1,2), (2,1). (2,1)
No. of favourable outcomes = 4
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 3)=$\dfrac{4}{{36}}{\text{ = }}\dfrac{1}{9}$
=364=91
(3) Sum of 4
a favourable outcome for Sum of 4 will be (2,2), (2,2), (3,1). (1,3)
No. of favourable outcomes = 4
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 4)=$\dfrac{4}{{36}}{\text{ = }}\dfrac{1}{9}$
(4) Sum of 5
favourable outcome for Sum of 5 will be (1,4), (4,1,), (2,3)., (2,3) (3,2). (3,2)
No. of favourable outcomes = 6
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 5)=$\dfrac{6}{{36}}{\text{ = }}\dfrac{1}{6}$
(5) Sum of 6
favourable outcome for Sum of 6 will be (1,5), (1,5), (2,4)., (2,4) (3,3). (3,3)
No. of favourable outcomes = 6
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 6)=$\dfrac{6}{{36}}{\text{ = }}\dfrac{1}{6}$
(6) Sum of 7
favourable outcome for Sum of 7 will be (1,6), (1,6), (2,5)., (2,5) (3,4). (3,4)
No. of favourable outcomes = 6
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 7)=$\dfrac{6}{{36}}{\text{ = }}\dfrac{1}{6}$
(7) Sum of 8
favourable outcome for Sum of 8 will be (2,6), (2,6,), (3,5). (3,5)
No. of favourable outcomes = 4
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 8)=$\dfrac{4}{{36}}{\text{ = }}\dfrac{1}{9}$
(8) Sum of 9
favourable outcome for Sum of 9 will be (6,3,), (3,6).
No. of favourable outcomes = 2
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 9)=$\dfrac{2}{{36}}{\text{ = }}\dfrac{1}{{18}}$
NOTE- In Such Types of Question first find out the total numbers of possible outcomes then find out the number of favourable cases, then divide them using the formula which stated above, we will get the required answer.
Complete step-by-step answer:
Here we have one unbiased dice which is numbered from 1 to 6. While other dice are biased on which numbers are 1,1,2,2,3,3.
When we through both the dice then every number of each dice will make a pair with each number of other dice so total numbers of possible outcomes will be
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
So total Favourable outcomes =36
The probability of getting each sum from 2 to 9
(1) Sum of 2
favourable outcome for Sum of 2 will be (1,1,), (1,1).
No. of favourable outcomes = 2
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 2)=$\dfrac{2}{{36}}{\text{ = }}\dfrac{1}{{18}}$
(2) Sum of 3
a favourable outcome for Sum of 3 will be (1,2), (1,2), (2,1). (2,1)
No. of favourable outcomes = 4
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 3)=$\dfrac{4}{{36}}{\text{ = }}\dfrac{1}{9}$
=364=91
(3) Sum of 4
a favourable outcome for Sum of 4 will be (2,2), (2,2), (3,1). (1,3)
No. of favourable outcomes = 4
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 4)=$\dfrac{4}{{36}}{\text{ = }}\dfrac{1}{9}$
(4) Sum of 5
favourable outcome for Sum of 5 will be (1,4), (4,1,), (2,3)., (2,3) (3,2). (3,2)
No. of favourable outcomes = 6
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 5)=$\dfrac{6}{{36}}{\text{ = }}\dfrac{1}{6}$
(5) Sum of 6
favourable outcome for Sum of 6 will be (1,5), (1,5), (2,4)., (2,4) (3,3). (3,3)
No. of favourable outcomes = 6
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 6)=$\dfrac{6}{{36}}{\text{ = }}\dfrac{1}{6}$
(6) Sum of 7
favourable outcome for Sum of 7 will be (1,6), (1,6), (2,5)., (2,5) (3,4). (3,4)
No. of favourable outcomes = 6
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 7)=$\dfrac{6}{{36}}{\text{ = }}\dfrac{1}{6}$
(7) Sum of 8
favourable outcome for Sum of 8 will be (2,6), (2,6,), (3,5). (3,5)
No. of favourable outcomes = 4
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 8)=$\dfrac{4}{{36}}{\text{ = }}\dfrac{1}{9}$
(8) Sum of 9
favourable outcome for Sum of 9 will be (6,3,), (3,6).
No. of favourable outcomes = 2
total number of possible cases = 36
As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (Sum of 9)=$\dfrac{2}{{36}}{\text{ = }}\dfrac{1}{{18}}$
NOTE- In Such Types of Question first find out the total numbers of possible outcomes then find out the number of favourable cases, then divide them using the formula which stated above, we will get the required answer.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Define the term system surroundings open system closed class 11 chemistry CBSE
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE