Probability
Aptitude

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Q.

A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that $|X| < 2$

 A.

5/7

 B.

3/7

 C.

3/5

 D.

1/3

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Solution:
Option(B) is correct

$|X|$ can take 7 values.
To get $|X| < 2$ ( i.e., $-2 < X < +2$) take $X= \{-1, 0, 1\}$

⇒ $P(|X| < 2) = \dfrac{\text{ Favourable Cases}}{\text{Total Cases}}$
= 3/7


(8) Comment(s)


Manchukonda Manoj Kumar
 ()

Numbers less than $|X| < 2$ means -3 & -2 also should be considerd, then why you have not considered, can anyone explain?


Riya
 ()

You are mistaken, $X<2$ will incorporate numbers -3 and -2, BUT $|X|<2$ means, $-2<X<2$. 
This, $|x|$ is called modulus of the number $X$.


JarredPef
 ()

Why people are saying there is an error?
I don't see anything wrong with the question.



Deepak
 ()

So out of $-3, -2, -1, 0, 1, 2, 3$, cases $-1,0,1$ are favorable ones. Which gives using the probability theorem,

$$P(|X| < 2) = \dfrac{\text{ Favourable Cases}}{\text{Total Cases}}$$

$$=\dfrac{3}{7}$$



Krugger
 ()

SmileSmile the answer is correct but there is also another easier method compared to that one.



Junaid Kazi
 ()

freely must explain then whole common person are understood.



Abhinav
 ()

in the given solution $-2<X>2$ seems to be wrong..$|X|<2$ implies $-2<X<2$


Shanky
 ()

Yes abhinav you are right, I also think there is a typo in line 2 of the solution.

I guess concerned people will recognize it make correction as needed. Smile