A box contains 5 red and 4 white marbles. Two marbles are drawn successively from the box without replacement and the second drawn marble is found to be white. Probability that the first is also white is?
$
  A.\dfrac{3}{8} \\
  B.\dfrac{1}{2} \\
  C.\dfrac{1}{3} \\
  D.\dfrac{1}{4} \\
 $

Hint: Here this is a certain and sure event. Which we will denote by S. This event has a certain number of samples. So, we will first have to calculate and write the samples and then find the probability of each event.

Complete step-by-step answer:
5 red and 4 white marbles are presented by ${R_1},{R_2},{R_3},{R_4},{R_5}$ and ${W_1},{W_2},{W_3},{W_4}$.
Box has total 5 red + 4 white =9 balls
Here, two marbles are drawn without the replacement.
To find the probability that the first marble is also white, if the second marble was found to be white.
As, the second ball drawn is white. So, the white ball left in the box will be 3. And also the total balls left will be 8.
Hence the probability that the first marble is also white will be:
$\therefore $Probability of white event = $\dfrac{{no.ofsampleSpaces}}{{TotalNo.ofSpaces}}$
=$\dfrac{3}{8}$
Thus required probability will be $\dfrac{3}{8}$.

So, the correct answer is “Option A”.

Note: Remember the concept about the sample space. Independent events, sure events, impossible events etc. Also, the concept of Permutation & Combination is also needed to be used in a proper way. As in above problem selection is applicable. So, combination rules were applicable.