Question

A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is

• A. $\frac{1}{20}$
• B. $\frac{1}{12}$
• C. $\frac{3}{10}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Given, a box contains 4 red balls & 6 black balls.
Let $P\left(R\right)$=Probability of red ball selection
$P\left(B\right)$ =Probability of black ball selection
Then, probability that three balls selected contains are red ball & two black balls, without replacement is
$P=P\left(RBB\right)+P\left(BRB\right)+P\left(BBR\right)$
$=P\left(R\right)P\left(B\right)P\left(B\right)+P\left(B\right)P\left(R\right)P\left(B\right)+P\left(B\right)P\left(B\right)P\left(R\right)$

$=\frac{4}{10}\frac{6}{9}\frac{5}{8}+\frac{6}{10}\frac{4}{9}\frac{5}{8}+\frac{6}{10}\frac{5}{9}\frac{4}{8}$
$=\frac{1}{2}$