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Q22(i):

Refer to Example 13. (i) Complete the following table: 

Solution :

Given: A table representing the sum of two dice thrown simultaneously, where the possible outcomes for the sum range from $2$ to $12$. The table provides the probability for some sums, but leaves the probabilities for sums $2, 3, 4, 5, 6, 7, 8, 9, 10, 11,$ and $12$ to be calculated.

To Find: Complete the probability distribution table for the sum of two dice.

Contextual Analysis: When two fair dice are thrown, each die has $6$ faces. The total number of possible outcomes is $6 \times 6 = 36$.

Sum on 2 dice 2 3 4 5 6 7 8 9 10 11 12
Probability $P(2)$ $P(3)$ $P(4)$ $P(5)$ $P(6)$ $P(7)$ $P(8)$ $P(9)$ $P(10)$ $P(11)$ $P(12)$

Step 1: Determine the number of favorable outcomes for each sum.
Let $(d_1, d_2)$ represent the outcome of the first and second die respectively. The total outcomes $n(S) = 36$.

  • Sum = 2: (1,1) $\rightarrow$ 1 outcome
  • Sum = 3: (1,2), (2,1) $\rightarrow$ 2 outcomes
  • Sum = 4: (1,3), (2,2), (3,1) $\rightarrow$ 3 outcomes
  • Sum = 5: (1,4), (2,3), (3,2), (4,1) $\rightarrow$ 4 outcomes
  • Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) $\rightarrow$ 5 outcomes
  • Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) $\rightarrow$ 6 outcomes
  • Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) $\rightarrow$ 5 outcomes
  • Sum = 9: (3,6), (4,5), (5,4), (6,3) $\rightarrow$ 4 outcomes
  • Sum = 10: (4,6), (5,5), (6,4) $\rightarrow$ 3 outcomes
  • Sum = 11: (5,6), (6,5) $\rightarrow$ 2 outcomes
  • Sum = 12: (6,6) $\rightarrow$ 1 outcome

Step 2: Calculate the probability for each sum.
Using the formula $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$, where total outcomes = $36$.

Sum 2 3 4 5 6 7 8 9 10 11 12
Probability $\frac{1}{36}$ $\frac{2}{36}$ $\frac{3}{36}$ $\frac{4}{36}$ $\frac{5}{36}$ $\frac{6}{36}$ $\frac{5}{36}$ $\frac{4}{36}$ $\frac{3}{36}$ $\frac{2}{36}$ $\frac{1}{36}$

Step 3: Simplify the fractions where applicable.

  • $P(2) = \frac{1}{36}$
  • $P(3) = \frac{2}{36} = \frac{1}{18}$
  • $P(4) = \frac{3}{36} = \frac{1}{12}$
  • $P(5) = \frac{4}{36} = \frac{1}{9}$
  • $P(6) = \frac{5}{36}$
  • $P(7) = \frac{6}{36} = \frac{1}{6}$
  • $P(8) = \frac{5}{36}$
  • $P(9) = \frac{4}{36} = \frac{1}{9}$
  • $P(10) = \frac{3}{36} = \frac{1}{12}$
  • $P(11) = \frac{2}{36} = \frac{1}{18}$
  • $P(12) = \frac{1}{36}$

Final Answer: The completed probability table is as follows: Sum 2: 1/36, Sum 3: 1/18, Sum 4: 1/12, Sum 5: 1/9, Sum 6: 5/36, Sum 7: 1/6, Sum 8: 5/36, Sum 9: 1/9, Sum 10: 1/12, Sum 11: 1/18, Sum 12: 1/36.


More Questions from Class 10 Mathematics Probability EXERCISE 14.1


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