# 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.

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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.

Last updated date: 24th Sep 2023

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