Question

Two balls are drawn one after another (without replacement) from a bag containing $2$ white, $3$ red and $5$ blue balls. What is the probability that atleast one ball is red?

• A. $\frac{7}{15}$
• B. $\frac{8}{15}$
• C. $\frac{7}{16}$
• D. $\frac{5}{16}$

Answer 1

Answer: (B)

$P$(atleast one red ball) $= 1 - P$(no red ball) Now to calculate $P$(no red ball), let $A$ be the event of drawing a non-red ball in first draw and $B$ be the event of drawing a non-red ball in second draw.
$\therefore P(A) = \frac{7}{10}$
and $P(B | A) = \frac{6}{9} = \frac{2}{3}$
$P$(no red ball) $= P(A \cap B) = P(A). P(B|A)$
$= \frac{7}{10}. \frac{2}{3}= \frac{7}{15}$
$\therefore P$(atleast one red ball) $=1- \frac{7}{15}= \frac{8}{15}$