Question

(i) Complete the following table:

(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 \(\frac{1}{11}\). Do you agree with this argument ? Justify your

Answer 1

(i) Completing the table: 

Possibility of throwing 2 dice with six faces, n(S) = 6 × 6 = 36. 

1. Sum on 2 dice will be 3 means (1,2), (2, 1) 

∴ n(A) = 2

2. Sum on 2 dice, getting 4 means 

(1, 3), (2, 2), (3, 1) 

∴ n(B) = 3

3. Sum of 2 dice, getting 5 means (1, 4), (2, 3), (3, 2), (4, 1) 

∴ n(C) = 4

4. Sum of 2 dice, getting 6 means (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) 

∴ n(D) = 5

5. Sum of 2 dice, getting 7 means (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) 

∴ n(E) = 6

6. Sum of 2 dice, getting 9 means (3, 6), (4, 5), (5, 4), (6, 3) 

∴ n(F) = 4

7. Sum of 2 dice, getting 10 means (4, 6), (5, 5), (6, 4) 

∴ n(G) = 3

8. Sum of 2 dice, getting sum 11 means (5, 6), (6, 5) 

∴ n(H) = 2

∴ The Table is completed as follows:

(ii) No. 11 sums are not having equal possibility. Here each sums are not event having equal possibility.