Two dice are rolled simultaneously and counts are added
(i) complete the table given below:
Event : 'Sum on 2 dice' 2 3 4 5 6 7 8 9 10 11 12
Probability 1/36 - - - - - 5/36 - - - 12/36
(ii) A student argues that 'there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.
Answers
Answer:
Probability of sum of dice in table below
Step-by-step explanation:
Two dice are rolled simultaneously => Total case = 6 * 6 = 36
Sum = 2 1+1 = 2 :only 1 case
Sum = 3 1+2 = 2 + 2 :2 Cases
Sum = 4 1+ 3 , 2 + 2 , 3 + 3 :3Cases
Sum = 5 1+4 , 2 + 3 , 3 + 2 , 4 + 1 :4 Cases
Sum = 6 1+5 ,2 + 4 , 3 + 3 , 4 + 2 , 5 + 1 : 5 Cases
Sum = 7 1+6 , 2+ 5 , 3 + 4 , 4 + 3 , 5+2 , 6 + 1 : 6 Cases
Sum = 8 2+6 , 3+ 5 , 4 + 4 , 5 + 3 , 6+2 : 5 Cases
Sum = 9 3+6 , 4+ 5 , 5 + 4 , 6 + 3 : 4 Cases
Sum = 10 4 + 6 , 5 + 5 , 6 + 4 : 3 cases
Sum = 11 5 + 6 , 6 + 5 : 2 Cases
Sum = 12 6 + 6 : 1 Case
Sum on Dice Probability
2 1/36
3 2/36 = 1/18
4 3/36 = 1/12
5 4/36 = 1/9
6 5/36
7 6/36 = 1/6
8 5/36
9 4/36 = 1/9
10 3/36 = 1/12
11 2/36 = 1/18
12 1/36
I do not agree with the argument that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11.
As probable number of outcomes are not equal
for Sum = 2 there is only 1 possible 1 & 1
while for Sum = 7 Max possibles 6 , 1+6 , 2+ 5 , 3 + 4 , 4 + 3 , 5+2 , 6 + 1
so Different probability
I think this helps for you
ok
