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 1 11 1 . Do you agree with this argument ? Justify your Select the correct answer from above options

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.