Question

There are 6 purple balls, 5 blue balls, and 3 green balls in a box. 5 balls were randomly chosen (without replacing them). Find the probability that

(a) Exactly 3 blue balls were chosen.

(b) 2 purple balls, 1 blue ball, and 2 green balls were chosen.

Answer 1

(a)

Blue = 5

Non-blue = 9

Total balls = 14

Number of ways of selecting 5 balls from 14 balls = 14C5 = 2002

Number of ways of selecting 3 blue balls from 5 blue balls = 5C3 = 10

Number of ways of selecting 2 Non- blue balls from 9 Non - blue balls = 9C2 = 36

So,

P( Exactly 3 blue balls were chosen. ) = 10 X 36/2002

                                                          = 0.1798

So,

Answer is:

0.1798

(b)

Purple = 6

Blue = 5

Green = 3

Total balls = 14

Number of ways of selecting 5 balls from 14 balls = 14C5 = 2002

Number of ways of selecting 2 Purple balls from 6 Purple balls = 6C2 = 15

Number of ways of selecting 1 Blue ball from 5 Blue balls = 5C1 = 5

Number of ways of selecting 2 Green balls from 3 Green balls = 3C2 = 3

So,

P( 2 purple balls, 1 blue ball, and 2 green balls were chosen) = 15 X 5 X 3/2002

                                                                                     = 0.1124

So,

Answer is:

0.1124