Question 955509
Let's find the probability that no balls are red.
Then the complement will have at least one red ball.
{{{P(NR1)=7/12}}}
{{{P(NR2)=6/12}}}
{{{P(NR3)=5/12}}}
So then,
{{{P(NR)=(7/12)(6/12)(5/12)}}}
{{{P(NR)=35/288}}}
So then,
{{{P=1-P(NR)}}}
{{{P=288/288-35/288}}}
{{{P=253/288}}}