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Mathematics

A dice has 6 faces marked by the given numbers as shown below :

123123\boxed{1} \quad \boxed{2} \quad \boxed{3} \quad \boxed{-1} \quad \boxed{-2} \quad \boxed{-3}

The dice is thrown once. What is the probability of getting

(i) a positive integer ?

(ii) an integer greater than -3 ?

(iii) the smallest integer ?

Probability

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Answer

Since, there are 6 faces in a dice.

∴ No. of possible outcomes = 6.

(i) Favourable outcomes for getting a positive integer are 1, 2, 3.

∴ No. of favourable outcomes = 3.

P(getting a positive integer) = No. of favourable outcomesNo. of possible outcomes=36=12\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.

Hence, the probability of getting a positive integer = 12\dfrac{1}{2}.

(ii) Favourable outcomes for getting an integer greater than -3 are -2, -1, 1, 2, 3.

∴ No. of favourable outcomes = 5.

P(getting an integer greater than -3)

= No. of favourable outcomesNo. of possible outcomes=56\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{6}.

Hence, the probability of getting an integer greater than -3 = 56\dfrac{5}{6}.

(iii) Favourable outcomes for getting smallest integer is -3.

∴ No. of favourable outcomes = 1.

P(getting smallest integer) = No. of favourable outcomesNo. of possible outcomes=16\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{6}.

Hence, the probability of getting smallest integer = 16\dfrac{1}{6}.

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