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Mathematics

A die is tossed once, the probability of getting a number not divisible by 2 or 3 is :

  1. 23\dfrac{2}{3}

  2. 56\dfrac{5}{6}

  3. 16\dfrac{1}{6}

  4. 13\dfrac{1}{3}

Probability

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Answer

On tossing a die, possible outcomes are {1, 2, 3, 4, 5, 6}.

No. of possible outcomes = 6

Favourable outcomes for getting a number not divisible by 2 or 3 are {1, 5}.

No. of favourable outcomes = 2

P(getting a number not divisible by 2 or 3) = No. of favourable outcomesNo. of possible outcomes=26=13\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}.

Hence, Option 4 is the correct option.

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