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

$\frac{7}{12}$

Let A be the event of not getting a red ball in f draw and B be the event of not getting a red ball in sec
draw.

∴ Required probability

= Probability that at least one ball is red

=1-Probability that none is red

= $1- P(A $ and $B)= 1 - P(A ∩ B)$

$= 1 - P(A) P(B/A) = 1- \frac{2}{3}× \frac{5}{8}=\frac{7}{12}$