Aptitude Discussion

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 |

**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**

**Manchukonda Manoj Kumar**

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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**

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Why people are saying there is an error?

I don't see anything wrong with the question.

**Deepak**

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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**

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

**Junaid Kazi**

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freely must explain then whole common person are understood.

**Abhinav**

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

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.

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