The probability of drawing exactly one red ball is given by:
[tex]P(1\ red\ ball)=\frac{4C1\times4C1}{8C2}=\frac{16}{28}[/tex]
the probability of drawing two red balls is given by:
[tex]P(2\ red\ balls)=\frac{4C2\times4C0}{8C2}=\frac{6}{28}[/tex]
The probability of drawing at least one red ball is:
P(1 red ball) + P(2 red balls) = 16/28 + 6/28 = 11/14.
The answer is [tex]\ \frac{11}{14}[/tex]
