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Mathematics

A number is chosen randomly from 10 to 99 (both inclusive) such that each number is equally likely to be chosen. The probability that at least one digit of the chosen number is 8 is:

  1. (15)\Big(\dfrac{1}{5}\Big)

  2. (19)\Big(\dfrac{1}{9}\Big)

  3. (110)\Big(\dfrac{1}{10}\Big)

  4. (1920)\Big(\dfrac{19}{20}\Big)

Probability

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Answer

The numbers are chosen from 10 to 99 inclusive.

Total number of outcomes = 90

Let E be the event of choosing a number with At Least One Digit as 8 , then

E = {18, 28, 38, 48, 58, 68, 78, 88, 98, 80, 81, 82, 83, 84, 85, 86, 87, 89}

The number of favorable outcomes to the event E = 18

∴ P(E) = Number of favorable outcomesTotal number of outcomes=1890=15\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{18}{90} = \dfrac{1}{5}

Hence, option 1 is the correct option.

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