Question

Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls one by one without replacement. What is the probability that at least one ball is red?

Answer 1

Let $A$ be the event of getting non-red ball in $1^{st}$ draw

Let $B$ be the event of getting non-red ball in $2^{nd}$ draw

$P\left(A\right)=\frac{6}{9}=\frac{2}{3}$

When a non-red ball is drawn in $1^{st}$ draw, there are $5$ non-red

balls and $3$ red balls.

$\therefore$ from $5$ non-red balls, one ball can be selected in ${}^5{C}_1$ ways

$\therefore P\left(B/A\right)=\frac{5}{8}$

Let $C$ be event that atleast one ball is red.

$P\left(C\right)=1-$ Probability of getting non-red ball

$=1-P(A\cap B)$

$=1-P\left(A\right).P\left(B/A\right)=1-\frac{2}{3}\times \frac{5}{8}$

$P\left(C\right)=\frac{7}{12}$