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Three balls are withdrawn (without replacement) from a bag containing 3 red, 4 green, and 5 blue balls.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
 #1
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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

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