Two balls are drawn from an urn containing 7 white, 6 red and 8 black balls one after the other without…
Two balls are drawn from an urn containing 7 white, 6 red and 8 black balls one after the other without replacement. Then the probability that atleast one of them is white, is
$\frac{4}{9}$
$\frac{13}{30}$
$\frac{11}{30}$
$\frac{17}{30}$
Solution
Given that urn contain 7 white, 6 red, 8 black balls.
Two balls are drawn one after the other without replacement, then following condition are there to get at least one of them is white
(a) First White, second non-white
$
\text { So, required probability }=\frac{7}{21} \times \frac{14}{20}
$
(b) First non-white, second white
$
\text { So, required probability }=\frac{14}{21} \times \frac{7}{20}
$
(c) Both are white $=\frac{7}{21} \times \frac{6}{20}$
$\therefore$ Required probability is
$
=\frac{7}{21} \times \frac{14}{20}+\frac{14}{21} \times \frac{7}{20}+\frac{7}{21} \times \frac{6}{20}=\frac{17}{30} \text {. }
$