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Probability — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsProbability2 Marks
Question
  1. Two dice, one blue and one grey, are thrown at the same time. Complete the following table:
    Event: Sum on 2 dice 2 3 4 5 6 7 8 9 10 11 12
    Probability $\frac{1}{{36}}$ $\frac{5}{{36}}$ $\frac{1}{{36}}$
  2. 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.
✓

Answer

  1. It can be observed that,
    To get the sum as 2, possible outcomes = (1, 1)
    To get the sum as 3, possible outcomes = (2, 1) and (1, 2)
    To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2)
    To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2)
    To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2), (3, 3)
    To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2), (3, 4), (4, 3)
    To get the sum as 8, possible outcomes = (6, 2), (2, 6), (3, 5), (5, 3), (4, 4)
    To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4)
    To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
    To get the sum as 11, possible outcomes = (5, 6), (6, 5)
    To get the sum as 12, possible outcomes = (6, 6)
    Event 2 3 4 5 6 7 8 9 10 11 12
    Probability $\frac{1}{{36}}$ $\frac{2}{{36}}$ $\frac{3}{{36}}$ $\frac{4}{{36}}$ $\frac{5}{{36}}$ $\frac{6}{{36}}$ $\frac{5}{{36}}$ $\frac{4}{{36}}$ $\frac{3}{{36}}$ $\frac{2}{{36}}$ $\frac{1}{{36}}$
  2. The probability of each of these sums will not be $\frac{1}{11}$ as their sums are not equally likely.

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