Probability
Aptitude

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Q.

A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either 3 or 13?

 A.

17/30

 B.

2/5

 C.

11/30

 D.

4/15

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Solution:
Option(B) is correct

We know that,

$\text{Probability} = \dfrac{\text{Favorable Cases}}{\text{Total Cases}}$

The probability that the number is a multiple of 3 is \(\dfrac{10}{30}\).

Since favorable cases here $\{3,6,9,12,15,18,21,24,27,30\}$

$=10 \text{ cases}$

Total cases= $30 \text{ cases}$

Similarly the probability that the number is a multiple of 13 is \(\dfrac{2}{30}\).

Since favorable cases here $\{13,26\}$

$=2 \text{ cases}$

Total cases= $30 \text{ cases}$

Neither 3 nor 13 has common multiple from 1 to 30. 

Hence, these events are mutually exclusive events. 

Therefore chance that the selected number is a multiple of 3 or 13 is:

\(=\dfrac{10+2}{30}\)

\(=\dfrac{2}{5}\)


(7) Comment(s)


Ayush
 ()

Answer and explaination given are write

But the option is wrong


Deepak
 ()

Thank you Ayush for letting me know the anomaly. Corrected it.


Ritika
 ()

Nicely explained. You people are doing a great job.



Who Gives
 ()

scared .... shocked.......unsure



Silpa
 ()

I think the option is wrong. Isn't it supposed to be (b)


Revant
 ()

Hi silpa,

You are right about the option.

They should put something(feedback for each question) so that we can tell them improve these bugs.