A die is thrown at random. Find the probability of getting

Question:

A die is thrown at random. Find the probability of getting

(i) 2

(ii) a number less than 3

(iii) a composite number

(iv) a number not less than 4.

Solution:

The possible outcomes when a dice is thrown at random are $1,2,3,4,5$ and 6 .

Total number of outcomes $=6$

(i) $\therefore \mathrm{P}_{\text {(getting 2) }}=\frac{1}{6}$

(ii) The numbers less than 3 are 1 and $2 .$

Number of possible outcomes $=2$

$\therefore \mathbf{P}_{\text {(getting a number }<3 \text { ) }}=\frac{2}{6}=\frac{1}{3}$

(iii) A composite number is defined as a number with at least one positive divisor other than itself and unity. In a dice, 4 and 6 are composite numbers.

Number of possible outcomes $=2$

$\therefore \mathrm{P}_{\text {(getting a composite number) }}=\frac{2}{6}=\frac{1}{3}$

(iv) A number not less than 4 can includes 4,5 or 6 .

Number of outcomes $=3$

$\therefore \mathrm{P}_{\text {(getting a number not less than 4) }}=\frac{3}{6}=\frac{1}{2}$

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