+256 Three balls are withdrawn (without replacement) from a bag containing 3 red, 4 green, and 5 blue balls. Find the probability that they are...
all the same color ?
all different colors?
+526 Total no. of balls \(=3+4+5\)
\(=12\)
3 balls are drawn without replacement.
⇒ No. of ways of choosing 3 balls \(=\binom{12}{3}\)
\(=220\)
Case 1: All are same color.
Favourable outcomes \(=\binom{3}{3}+\binom{4}{3}+\binom{5}{3}\)
\(=1+4+10\)
\(=15\)
∴ P(same color balls) \(= {15\over 220}\)
\(={3 \over 44}\)
Thus, the probability that they're same color is \(3 \over 44\).
Case 2: All are different colors.
Favourable outcomes \(=\binom{3}{1}*\binom{4}{1}*\binom{5}{1}\)
\(=3*4*5\)
\(=60\)
∴ P(different color balls) \(={60 \over 220}\)
\(={3 \over 11}\)
Thus, the probability that they're of different colors is \(3 \over 11\).
