Question

We have 4 different colors of balls available. 4 are of red, 3 are green, 2 are blue and 7 are of black color. In how many ways can a child take one blue ball, one red ball and one ball of either green or black ball?

Answer 1

Answer:

52 ways

Step-by-step explanation:

Total balls = 4 + 3 + 2 + 7 = 16

P of blue ball = 2/16 = 1/8

P of redball = 4/15

P of either black or green = (3+7)/14 = 10/14

By Product rule,

P of one blue, then one red and one ball of either black or green = 2/16 * 4/15 * 10/14 = 1/52 [ By reducing

the fractions]

So total there are 52 ways of doing so

Answer 2

Answer:

80 possible combinations.

Step-by-step explanation:

for choosing 1 blue ball out of 2 blue balls

=2c1 = 2

for choosing 1 red ball out of 4 red balls

=4c1 = 4

for choosing 1 green or black ball out of a total of 10 balls (black + green)

= 10c1 = 10

therefore the total combination

= 2 x 4 x 10 = 80