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$

A.\dfrac{3}{8} \\

B.\dfrac{1}{2} \\

C.\dfrac{1}{3} \\

D.\dfrac{1}{4} \\

$

Answer

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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 $\dfrac{3}{8}$.