Question

: A box contains 3 white and 2 black balls. Two balls are drawn at random, one after the other. If the balls are not replaced. What is the probability that both the balls are black?
A.$\dfrac{2}{5}$
B.$\dfrac{1}{5}$
C.$\dfrac{1}{{10}}$
D.None of the above

Answer 1

Hint : Probability means the certainty of occurring of any event. To find the probability of an experiment for which the outcomes can’t be guessed with certainty (Random experiment), two definitions are there, one is an event, and the other is sample space.
Event is the other name of the favorable outcome of any experiment, while Sample space is the set of all possible outcomes of that experiment, and we can say that an event will be a subset of sample space.
Probability for any random experiment is given by:
${\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}} = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$

Complete step-by-step answer :
In this question, we need to determine the probability that the process of drawing the balls without replacement until all the 2 black balls come out of the bag.
Number of white balls = 3
Number of black balls = 2
Hence the total number of balls= 3+2 =5
Hence the probability of drawing a black ball in the first draw is given as:

$${P_1} = \dfrac{{{}^2{C_1}}}{{{}^5{C_1}}} \\\ = \dfrac{{\left( {\dfrac{{2!}}{{1!}}} \right)}}{{\left( {\dfrac{{5!}}{{4! \times 1!}}} \right)}} \\\ = \dfrac{2}{5} \\\$$

Now the box is left with only one black ball and a total of 3+1=4 balls
Hence the probability of drawing a second black ball is given as:

$${P_2} = \dfrac{{{}^1{C_1}}}{{{}^4{C_1}}} \\\ = \dfrac{{\left( {\dfrac{{1!}}{{0!}}} \right)}}{{\left( {\dfrac{{4!}}{{3! \times 1!}}} \right)}}{\text{ }}\left[ {0! = 1} \right] \\\ = \dfrac{1}{4} \\\$$

The probability that both balls are black= probability of the first ball being black X Probability of the second ball being black is given as:

$$P = {P_1} \times {P_2} \\\ = \dfrac{2}{5} \times \dfrac{1}{4} \\\ = \dfrac{1}{{10}} \\\$$

Hence the probability of drawing both balls black is $\dfrac{1}{{10}}$
So, the correct answer is “Option C”.

Note : Students should be careful in this type of question as it is mentioned that the ball is not replaced as the number of balls reduces after every draw, but if the balls are replaced, then the number of balls remains the same.