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# A number is chosen at random from the numbers$- 3, - 2, - 1,0,1,2,3$. What will be the probability that the square of this number is less than or equal to $1$?A) $\dfrac{1}{7}$ B) $\dfrac{2}{7}$ C) $\dfrac{3}{7}$ D) $\dfrac{4}{7}$  Verified
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Hint: Only the square of numbers $1,0{\text{ and - 1}}$ will be less than or equal to 1.So the favourable outcomes are 3 out of 7 total outcomes.

Given, total numbers=$\left\{ { - 3, - 2, - 1,0,1,2,3} \right\}$ Then total number of outcomes(n) are $7$ .The squares of these numbers =$\left\{ {9,4,1,0,1,4,9} \right\}$.Now we have to find the probability of choosing the number whose square will be less than or equal to zero. Let the event of choosing the number whose square is less than or equal to one be E then E$= \left\{ { - 1,0,1} \right\} = 3$ favourable outcomes because only the squares of $1,0{\text{ and - 1}}$ will be less than or equal to one.
Then , then probability P(E)=$\dfrac{{\text{E}}}{{\text{n}}}$
P(E)$= \dfrac{3}{7}$
$\Rightarrow {\text{Probability = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$ .
$\Rightarrow {\text{Probability = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Sample space}}}}$