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Considering an experiment of tossing two fair dice, one black and one red. What is the probability that the number on the black die divides the number on red die?

  1. $\dfrac{22}{36} \\$
  2. $\dfrac{12}{36} \\$
  3. $\dfrac{14}{36} \\$
  4. $\dfrac{6}{36}$
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If number on the black die divides the number on red die, respective numbers on the black die and red die must be any of the following

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), (2,2),(2,4),(2,6), (3,3),(3,6), (4,4), (5,5), (6,6)

Total number of possible outcomes = 14

The probability that the number of the black die divides the

number of red die = $\frac{14}{36}$

Hence,Option(c)$\frac{14}{36}$.

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1st die  (1,2,3,4,5,6): 6 ways
2nd die (1,2,3,4,5,6): 6 ways

Thus On tossing two dice, total number of possible outcomes = 6 x 6 = 36

  • If 1 comes in black dice, number on red dice divisible by 1 = {1,2,3,4,5,6}  // 6 success
  • If 2 comes in black dice, number on red dice divisible by 2 = {2,4,6}  // 3 success
  • If 3 comes in black dice, number on red dice divisible by 3 = {3,6}  // 2 success
  • If 4 comes in black dice, number on red dice divisible by 2 = {4}  // 1 success
  • If 5 comes in black dice, number on red dice divisible by 2 = {5}  // 1 success
  • If 6 comes in black dice, number on red dice divisible by 2 = {6}  // 1 success


Total number of favorable outcomes = 14

Required probability = 14/36

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