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# Probability Problem

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Three balls are withdrawn (without replacement) from a bag containing 3 red, 4 green, and 5 blue balls.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?

Apr 30, 2021

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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 $$3 \over 11$$.

Apr 30, 2021