Math, asked by suyash0701, 1 month ago

show the sample space when 2 unbiased dice are thrown also compare the probabilities of all possible sum of outcomes appear. Also prepare a detailed note including examples that why the probability of each possible sum of dice is not same.

Answers

Answered by amitnrw
2

Given  : 2 unbiased dice are thrown

To Find : sample space  

 probabilities of all possible sum of outcomes appear.

Solution:

unbiased dice has numbers from 1 to 6

two Dice are rolled hence total output =  6 * 6  = 36

hence n(S) = 36

Numbers of elements   in the sample space 36

S  =  {  (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) }

Possible Sum  are from 2 to  12

Sum      Cases                 Frequency           Probability      

2           { (1 , 1)}                            1                     1/36

3           { (1,2),(2,1)}                      2                    2/36 = 1/18

4           { (1,3),(2,2)(3,1)}              3                    3/36 = 1/12

5           { (1,4),(2,3),(3,2)(4,1)}      4                    4/36 = 1/9

6      { (1,5),(2,4),(3,3)(4,2)(5,1)}    5                    5/36

7  { (1,6),(2,5),(3,4)(4,3) ,(5,2),(6,1)}    6               6/36  = 1/6

8      { (2,6),(3,5),(4,4)(5,3)(6,2)}    5                    5/36

9           { (3,6),(4,5),(5,4)(6,3)}      4                    4/36 = 1/9

10           { (4,6),(5,5)(6,4)}              3                    3/36 = 1/12

11           { (5,6),(6,5)}                      2                    2/36 = 1/18

12           { (6 ,6)}                            1                      1/36

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