Probability
Aptitude

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Q.

A set of cards bearing the number 200-299 is used in a game. if a card is drawn at random, what is the probability that it is divisible by 3:

 A.

0.66

 B.

0.33

 C.

0.44

 D.

0.55

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Solution:
Option(B) is correct

Total numbers of cards = 100 {and not 99}

Multiples of three = 33

Edit: To check the detailed calculation on how multiple of three=33 is calculated, check comment by Shelly.

The required probability,

$=\dfrac{33}{100}$

$=\textbf{0.33}$

Edit 2: For a shortcut alternative solution, check comment by Sanajit Ghosh.


(6) Comment(s)


Sanajit Ghosh
 ()

Easiest way

$T_n=a+(n-1)d$ where 1st number divisible by 3 is $a=201$,$d=3$,$T_n=297$ we have to find $n$, i.e number div by 3

Solving, we get $n=33$,

Therefore, total no = 200 to 299 i.e 100 nos

So $P=\dfrac{33}{100}$



Shelly
 ()

Explanation on arriving at the count of multiple of three = 33

All the number divisible by 3 from an A.P. with a common difference, $d$ of 3 as follows.

$201,204, 207,.....297.$

1st term, $a=201$.

If the number of terms in the above A.P. is $n$, we know from the properties of an A.P.

$\text{last term}=a+(n-1)d$

$\Rightarrow 297=201+(n-1)\times3$

$\Rightarrow \dfrac{297-201}{3}=(n-1)$

$\Rightarrow \dfrac{96}{3}=(n-1)$

$\Rightarrow n=32+1$

$\Rightarrow n=\textbf{33}$



Tejaswini
 ()

multiples of three=33



Shivanand
 ()

There are a 100 cards totally. And out of those 100 cards, 33 are divisible by 3. So the answer must be 0.33



Steve
 ()

This question was just stupid didn't state what the author meant.



Satish Bhat
 ()

what is the probability that it is divisible by ?

and

Total numbers of cards = 100 not 200.