Multiple choice

Two symmetrical dice are thrown. The probability of throwing a doublet such that their sum is less than $9$ is

  1. $\dfrac{1}{36}$
  2. $\dfrac{1}{18}$
  3. $\dfrac{1}{9}$
  4. $\dfrac{1}{72}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total outcomes = 36. Doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). Sums are 2, 4, 6, 8, 10, 12. Sums < 9 are 2, 4, 6, 8. There are 4 such doublets. Probability = 4/36 = 1/9.

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