Introduction to pair of dice:
A die is a cube with the numbers 1 to 6 in each of its faces. This is utilized to play different games. When we throw a dice, we can expect any one of the six values { 1,2,3,4,5,6} to come as an output. Therefore there are 6 chances that can be expected when we throw a die.
Similarly if two dice are thrown, we can expect 6 `xx ` 6 = 36 chances. [Any one of the 36 chances can be expected.] The 36 chances are given as follows:
{ ( 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) }
By knowing all these 36 chances, we can evaluate the problems involving in it. Is this topic Percentage Difference Calculator hard for you? Watch out for my coming posts.
If we throw tow dice at a time, we can expect any one of 36 chances to come as an output.
Therefore to an output of any one of the 36 chances, the probability can be written as `1/36` .
The formula to fin the probability is given by:
Probability = `["Number of favourable cases"] / ["Total number of possible cases"]` .
Practice Problems on Pair of Dice.
Ex 1: The dice are thrown simultaneously, Find the probability of getting:
(i) an even number as the sum
(ii) the sum as a prime number.
Solution:
(i) Let A be the event of getting an even number as the sum that 2,4,6,8,10,12.
Elementary events favourable to event A are:
{( 1,1) ( 1,3) (3,1), ( 2,2) ( 1,5) (5,1), ( 2,4) (4,2) (3,3), (2,6), ( 6,2), (4,4), (5,3), (3,5) , (5,5), ( 6,4), (4,6), and ( 6,6)}.
Clearly, favourable number of elementary events = 18.
Hency, required probability = `18/36` = `1/2` .
(ii) Let A be the event of getting the sum as a prime number that 2,3,5,7,11.
Elementary events favourable to event A are: { ( 1,1),( 1,2), (2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2).(3.4). (4.3).(6,5) and (5,6)}.
Therefore Favourable number of elementary events =15.
Hence, required probability = `15/36` = `5/12` .
Ex 2: The dice are thrown simultaneously, Find the probability of getting:
(i) a total of at least 10
(ii) a doublet of even number.
Solution: (i)Let A be the event of getting a total of at least 10 that is 10, 11,12.
Then, the elementary events favourable to A are: {(6,4), (4,6), (5,5), (6,5), (5,6), (6,6)}.
Therefore Favrouable number of elementary events = 6.
Hence, required probability = `6/36` = `1/6` .
(ii) Let A be the event of getting a doublet of even number.
Then, the elementary events favourable to A are (2,2), (4,4) and (6,6)
Therefore Favrouable number of elementary events = 3.
Hence, required probability `3/36` = `1/12` .
Ex 3: The dice are thrown simultaneously, Find the probability of getting:
(i) a multiple of 2 on one dice and a multiple of 3 on the other.
(ii) same number on both dice that is a doublet.
(iii) a multiple of 3 as the sum.
Solution:
(i) Let A be the event of getting a multiple of 2 on one die and a multiple of 3 on the other.
Then, the elementary events favourable to A are: {(2,3),(2,6),(4,3),(4,6),(6,3),(6,6),(3,2),(3,4),(,6),(6,2),(6,4)}.
Therefore Favourable number of elementary events = 11.
Hence required probability = `11/36` .
(ii) Let A be the event of getting the same number on both dice.
Then elementary events favourable to A are: (1,1),(2,2), (3,3),(4,4),(5,5) and (6,6)
Therefore Favorable number of elementary events = 6.
Hence, required probability = `6/36` = `1/6` .
(iii) Let A be the event of getting a multiple of 3 as the sum that 3,6,9,12.
Then elementary events favourable to A are: {(1,2),(2,1),(1,5),(5,1,) (2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}.
Therefore Favourable number of elementary events = 12.
Hence, required probability = `12/36` = `1/3` .
I have recently faced lot of problem while learning free online college algebra solver, But thank to online resources of math which helped me to learn myself easily on net.
Practice Problems on Pair of Dice.
1. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
[ Ans: (1) `5/36` , (ii) `1/6` , ( iii) `1/12` , (iv) `1/12` ]
2. (i) a sum greater than 9
(ii) an even number on first
(iii) an even number on one and a multiple of 3 on the other
(iv) neither 9 nor 11 as the sum of the number on the faces.
[Ans: (i) `1/6` , (ii) `1/2` , (iii) `11/36` , (iv)` 5/6` ].
A die is a cube with the numbers 1 to 6 in each of its faces. This is utilized to play different games. When we throw a dice, we can expect any one of the six values { 1,2,3,4,5,6} to come as an output. Therefore there are 6 chances that can be expected when we throw a die.
Similarly if two dice are thrown, we can expect 6 `xx ` 6 = 36 chances. [Any one of the 36 chances can be expected.] The 36 chances are given as follows:
{ ( 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) }
By knowing all these 36 chances, we can evaluate the problems involving in it. Is this topic Percentage Difference Calculator hard for you? Watch out for my coming posts.
If we throw tow dice at a time, we can expect any one of 36 chances to come as an output.
Therefore to an output of any one of the 36 chances, the probability can be written as `1/36` .
The formula to fin the probability is given by:
Probability = `["Number of favourable cases"] / ["Total number of possible cases"]` .
Practice Problems on Pair of Dice.
Ex 1: The dice are thrown simultaneously, Find the probability of getting:
(i) an even number as the sum
(ii) the sum as a prime number.
Solution:
(i) Let A be the event of getting an even number as the sum that 2,4,6,8,10,12.
Elementary events favourable to event A are:
{( 1,1) ( 1,3) (3,1), ( 2,2) ( 1,5) (5,1), ( 2,4) (4,2) (3,3), (2,6), ( 6,2), (4,4), (5,3), (3,5) , (5,5), ( 6,4), (4,6), and ( 6,6)}.
Clearly, favourable number of elementary events = 18.
Hency, required probability = `18/36` = `1/2` .
(ii) Let A be the event of getting the sum as a prime number that 2,3,5,7,11.
Elementary events favourable to event A are: { ( 1,1),( 1,2), (2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2).(3.4). (4.3).(6,5) and (5,6)}.
Therefore Favourable number of elementary events =15.
Hence, required probability = `15/36` = `5/12` .
Ex 2: The dice are thrown simultaneously, Find the probability of getting:
(i) a total of at least 10
(ii) a doublet of even number.
Solution: (i)Let A be the event of getting a total of at least 10 that is 10, 11,12.
Then, the elementary events favourable to A are: {(6,4), (4,6), (5,5), (6,5), (5,6), (6,6)}.
Therefore Favrouable number of elementary events = 6.
Hence, required probability = `6/36` = `1/6` .
(ii) Let A be the event of getting a doublet of even number.
Then, the elementary events favourable to A are (2,2), (4,4) and (6,6)
Therefore Favrouable number of elementary events = 3.
Hence, required probability `3/36` = `1/12` .
Ex 3: The dice are thrown simultaneously, Find the probability of getting:
(i) a multiple of 2 on one dice and a multiple of 3 on the other.
(ii) same number on both dice that is a doublet.
(iii) a multiple of 3 as the sum.
Solution:
(i) Let A be the event of getting a multiple of 2 on one die and a multiple of 3 on the other.
Then, the elementary events favourable to A are: {(2,3),(2,6),(4,3),(4,6),(6,3),(6,6),(3,2),(3,4),(,6),(6,2),(6,4)}.
Therefore Favourable number of elementary events = 11.
Hence required probability = `11/36` .
(ii) Let A be the event of getting the same number on both dice.
Then elementary events favourable to A are: (1,1),(2,2), (3,3),(4,4),(5,5) and (6,6)
Therefore Favorable number of elementary events = 6.
Hence, required probability = `6/36` = `1/6` .
(iii) Let A be the event of getting a multiple of 3 as the sum that 3,6,9,12.
Then elementary events favourable to A are: {(1,2),(2,1),(1,5),(5,1,) (2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}.
Therefore Favourable number of elementary events = 12.
Hence, required probability = `12/36` = `1/3` .
I have recently faced lot of problem while learning free online college algebra solver, But thank to online resources of math which helped me to learn myself easily on net.
Practice Problems on Pair of Dice.
1. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
[ Ans: (1) `5/36` , (ii) `1/6` , ( iii) `1/12` , (iv) `1/12` ]
2. (i) a sum greater than 9
(ii) an even number on first
(iii) an even number on one and a multiple of 3 on the other
(iv) neither 9 nor 11 as the sum of the number on the faces.
[Ans: (i) `1/6` , (ii) `1/2` , (iii) `11/36` , (iv)` 5/6` ].
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