Question

A bowl contains 4 red, 5 blue and 6 white balls. If 4 balls are taken at random with replacement, find the probability of having

4 blue balls

4 white balls

2 red balls

3 blue balls

2 red, 1 blue, and 1 white balls

Answer 1

Solution:

Probabaility of having 4 blue balls = (4×4×4×4)/(15×15×15×15)

= 256/50625 = 0.00505

Solution(b)

Probability of choosing 4 white balls

= (6×6×6×6)/(15×15×15×15)

= 1296/50625

= 0.0256

Solution(c)

Probability of 2 red balls and 2 other color

=(4×4×11×11)/(15×15×15×15) = 1936÷50625 = 0.0382

Solution(d)

3 blue balls and 1 other type of color

(5×5×5×10)/(15×15×15×15) = 1250/50625 = 0.0247
Solution(e)

Probability of 2 red, 1 blue and 1 white balls

(4×4×5×6)/(15*15*15*15) = 480/50625 =0.00948