Question

An urn contains 8 black marbles and 4 white marbles. Two marbles are chosen at random and without replacement. Then the probability that both marbles are black is

• A. $7/33$
• B. $2/3$
• C. $7/11$
• D. $14/33$
• E. $21/143$

Answer 1

Answer: (D)

$14/33$

Answer 2

Given data:

Black marble $=8$

White marbles $=4$

Then,

Step 1

Probability of drawing a black marble on the first draw = \dfrac{(Number of black marbles)}{(Total number of marbles)}

$= \dfrac{8}{(8 + 4)} =\dfrac{ 2}{3}$

Step 2

After the first black marble is drawn, there will be $7$ black marbles left and $4$ white marbles, making a total of $11$ marbles for the second draw (since 1 marble has already been taken out).

Probability of drawing a black marble on the second draw, given that the first marble was black = (Number of remaining black marbles) / (Total number of remaining marbles)$= 7 / 11$.

Step 3 : Now the overall probability of drawing two black marbles

Probability of drawing two black marbles = Probability of first black marble × Probability of second black marble

$= (2/3) × (7/11) =14/33$

Hence the required answer is $14/33$.