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
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 n any dice.
(xiv) even number on each die
(xv) 5 as the sum
(xvi) 2 will come up at least once
(xvii) 2 will not come either time
Answers
In a simultaneous throw of a pair of dice, probability of getting:
(i) 8 as the sum =5/36
(ii) a doublet =1/6
(iii) a doublet of prime numbers =1/12
(iv) a doublet of odd numbers =1/12
(v) a sum greater than 9 =5/36
(vi) an even number on first = 1/2
(vii) an even number on one and a multiple of 3 on the other =11/36
(viii) neither 9 nor 11 as the sum of the numbers on the faces =5/6
(ix) a sum less than 6 =5/18
(x) a sum less than 7 =7/18
(xi) a sum more than 7 =11/18
(xii) at least once = data insufficient
(xiii) a number other than 5 in any dice.=25/36
(xiv) even number on each die=1/4
(xv) 5 as the sum=1/9
(xvi) 2 will come up at least once=11/36
(xvii) 2 will not come either time=25/36
Step 1:
When a pair of dice is thrown simultaneously total number of outcome is 6*6=36.
Step 2:
Problem (i) :
To get 8 as a sum outcomes are
(2,6),(6,2),(3,5),(5,3),(4,4)
Total number of outcome is 5
There the probability of getting 8 as sum is =5/36
Problem (ii) :
To get doublet the outcomes are
(1,1),(2,2),(3,3),(4,4),(5 5),(6,6)
Total number of outcome is 6.
Therefore probability of getting doublet is 6/36=1/6.
Problem (iii) :
To get the doublet of prime number the outcomes are (2,2),(3,3),(5,5)
Total number of outcome is 3.
Therefore probability of getting doublet of prime number is 3/36=1/12.
Problem (iv) :
To get doublet of odd number the outcomes are (1,1), (3,3),(5,5).
Total number of outcomes is 3 .
Therefore probability of getting doublet of odd numbers is 3/36=1/12.
Problem (v) :
To get the sum greater than 9 the outcomes are (4 ,6),(6,4),(5,6),(6,5), (6,6).
Total number of outcomes is 5.
Therefore probability of getting sum greater than 9 is =5/36.
Problem (vi) :
To get the even number on first dice the outcomes are (2,1),(2,2),(2,3), (2,4),(2,5),(2,6)..... 6 outcomes
Similar for considering 4and 6 at first dice we get 6 outcomes for each case.
Now total number of outcome is 6*3=18.
Therefore probability of getting even number on first dice is 18/36=1/2.
Problem (vii)
To get even number on one dice and multiple of 3 on other dice the outcomes are(2,3),(2,6),(3,2),(6,2),(4,3),(4,6),(3,4),(6,4),(6, 3),(6,6),(3,6)
Total number of outcome is 11.
Therefore probability of getting even number in one dice and multiple of odd number in other dice is 11/36.
Problem (vii) :
To get neither 9nor 11 as sum number of faces we have to consider other than the sum of 9 and 11 number on the faces.
So the outcomes of sum 9 and 11 are (3,6),(6,3),(4,5),(5,4),(5,6),(6,5).
Here total number of outcomes are 6.
Therefore probability of getting neither 9 nor 11 as sum on dice is 1-6/36=5/6.
Problem (ix)
To get the sum less than 6 the outcomes are (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1).
Total number of outcomes is 10.
Therefore probability of getting sum less than 6 is 10/36=5/18.
Problem (x) :
To get the sum less than 7 the outcomes are (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2)
Total number of outcomes is 14.
Therefore probability of getting sum less than 7 is =14/36=7/18.
Problem (xi) :
To get the sum more than 7 the outcomes are other (1,1),(1,2),(1,3),(1,4), (1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2)
Total number of outcomes is 36-14=22
Therefore probability of getting sum more than 7 is 22/36=11/18.
Problem (xiii)
To get the number other than 5 the outcomes are other than (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5)
Total number of outcomes is 36-11=25.
Therefore probability of getting number other than 5 is 25/36
Problem (xiv)
To get even number in each die the outcomes are (2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)
Total number of outcome is 9.
Therefore probability of getting even number in each dice is =9/36=1/4.
Problem (xv)
To get the sum as 5 the outcomes are (1,4),(2,3),(3,2),(4,1).
Total number of outcomes is 4
Therefore probability of getting sum as 5 = 4/36 =1/9
Problem (xvi)
2 will come up at least once here the outcomes are
(1,2),(2,2), (2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2).
Total number of outcome is 11
Therefore probability =11/36.
Problem (xvii)
In this problem we have to calculate the probability of 2 will not come either.
In the previous problem we have calculate 2 will come up at least once
Therefore here the probability is
1-11/36=25/36.
Answer:
In a simultaneous throw of a pair of dice, probability of getting:
(i) 8 as the sum =5/36
(ii) a doublet =1/6
(iii) a doublet of prime numbers =1/12
(iv) a doublet of odd numbers =1/12
(v) a sum greater than 9 =5/36
(vi) an even number on first = 1/2
(vii) an even number on one and a multiple of 3 on the other =11/36
(viii) neither 9 nor 11 as the sum of the numbers on the faces =5/6
(ix) a sum less than 6 =5/18
(x) a sum less than 7 =7/18
(xi) a sum more than 7 =11/18
(xii) at least once = data insufficient
(xiii) a number other than 5 in any dice.=25/36
(xiv) even number on each die=1/4
(xv) 5 as the sum=1/9
(xvi) 2 will come up at least once=11/36
(xvii) 2 will not come either time=25/36
Step 1:
When a pair of dice is thrown simultaneously total number of outcome is 6*6=36.
Step 2:
Problem (i) :
To get 8 as a sum outcomes are
(2,6),(6,2),(3,5),(5,3),(4,4)
Total number of outcome is 5
There the probability of getting 8 as sum is =5/36
Problem (ii) :
To get doublet the outcomes are
(1,1),(2,2),(3,3),(4,4),(5 5),(6,6)
Total number of outcome is 6.
Therefore probability of getting doublet is 6/36=1/6.
Problem (iii) :
To get the doublet of prime number the outcomes are (2,2),(3,3),(5,5)
Total number of outcome is 3.
Therefore probability of getting doublet of prime number is 3/36=1/12.
Problem (iv) :
To get doublet of odd number the outcomes are (1,1), (3,3),(5,5).
Total number of outcomes is 3 .
Therefore probability of getting doublet of odd numbers is 3/36=1/12.
Problem (v) :
To get the sum greater than 9 the outcomes are (4 ,6),(6,4),(5,6),(6,5), (6,6).
Total number of outcomes is 5.
Therefore probability of getting sum greater than 9 is =5/36.
Problem (vi) :
To get the even number on first dice the outcomes are (2,1),(2,2),(2,3), (2,4),(2,5),(2,6)..... 6 outcomes
Similar for considering 4and 6 at first dice we get 6 outcomes for each case.
Now total number of outcome is 6*3=18.
Therefore probability of getting even number on first dice is 18/36=1/2.
Problem (vii)
To get even number on one dice and multiple of 3 on other dice the outcomes are(2,3),(2,6),(3,2),(6,2),(4,3),(4,6),(3,4),(6,4),(6, 3),(6,6),(3,6)
Total number of outcome is 11.
Therefore probability of getting even number in one dice and multiple of odd number in other dice is 11/36.
Problem (vii) :
To get neither 9nor 11 as sum number of faces we have to consider other than the sum of 9 and 11 number on the faces.
So the outcomes of sum 9 and 11 are (3,6),(6,3),(4,5),(5,4),(5,6),(6,5).
Here total number of outcomes are 6.
Therefore probability of getting neither 9 nor 11 as sum on dice is 1-6/36=5/6.
Problem (ix)
To get the sum less than 6 the outcomes are (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1).
Total number of outcomes is 10.
Therefore probability of getting sum less than 6 is 10/36=5/18.
Problem (x) :
To get the sum less than 7 the outcomes are (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2)
Total number of outcomes is 14.
Therefore probability of getting sum less than 7 is =14/36=7/18.
Problem (xi) :
To get the sum more than 7 the outcomes are other (1,1),(1,2),(1,3),(1,4), (1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2)
Total number of outcomes is 36-14=22
Therefore probability of getting sum more than 7 is 22/36=11/18.
Problem (xiii)
To get the number other than 5 the outcomes are other than (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5)
Total number of outcomes is 36-11=25.
Therefore probability of getting number other than 5 is 25/36
Problem (xiv)
To get even number in each die the outcomes are (2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)
Total number of outcome is 9.
Therefore probability of getting even number in each dice is =9/36=1/4.
Problem (xv)
To get the sum as 5 the outcomes are (1,4),(2,3),(3,2),(4,1).
Total number of outcomes is 4
Therefore probability of getting sum as 5 = 4/36 =1/9
Problem (xvi)
2 will come up at least once here the outcomes are
(1,2),(2,2), (2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2).
Total number of outcome is 11
Therefore probability =11/36.
Problem (xvii)
In this problem we have to calculate the probability of 2 will not come either.
In the previous problem we have calculate 2 will come up at least once.
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