# Sam rolled two differenet colored 6 sided dice a total of 4 times. What is the probablity that he rolled a sum of 6 all 4 times and that the sum of 6 was acheived in a different way each time?

llltkl | Student

The pair of dice can come up in the following ways:

11  21  31  41  51  61

12  22  32  42  52  62

13  23  33  43  53  63

14  24  34  44  54  64

15  25  35  45  55  65

16  26  36  46  56  66

The probability that a sum of 6 will come up in first roll is 5/36.

In the second roll, a sum of 6 has to be rolled through a different combination.

The probability is 4/36.

Similarly, such probability in the third roll is: 3/36 and in the fourth roll, 2/36.

The events are dependent, hence overall probability is: 5/36*4/36*3/36*2/36

=0.0000714

aruv | Student

A pair of dice  is rolled 4 times.

Thus total number of possible out comes are =

E=Getting a sum of 6, its sample space is

{(1,5),(2,4),(3,3),(4,2),(5,1)}

Thus total possible number of favourable out comes=5

But the sum of 6 was to be acheived in a different way each time in 4 rolles. It will be

P(E)=`(^5P_4)/(36)^4`

`=(5xx4xx3xx2)/(36)^4`

`=.0000714`

aruv | Student

A pair of dice  is rolled 4 times.

Thus total number of possible out comes are =`(36)^4=1679616`

E=Getting a sum of 6, its sample space is

{(1,5),(2,4),(3,3),(4,2),(5,1)}

Thus total possible number of favourable out comes=5

Let

p=P(E)=5/36

Getting a sum of 6, all 4 times = `^4C_4(5/36)^4(31/36)^0`

`=(5/36)^4=.000372`

aruv | Student

Total number of possible outcomes in this experiment=`6^4=1296`

E=Event getting sum of 4={(1,1,1,3),(1,1,3,3),(1,3,1,1),(3,1,1,1),(1,1,2,2),(1,2,1,2),(2,1,1,2),(1,2,2,1),(2,1,2,1),(2,2,1,1)}

Thus total number of favourable outcomes=10

P(E)=10/1296

=.00772