Answer to Question #297625 in Statistics and Probability for Minion

Question

Combination

A bag of contains 4 yellow, 8 red and 3 blue balls. How many ways can 3 balls be selected so that the following conditions are met?

a. all balls are yellow

b. all balls are red

c. There is one ball from each color

d. there are 2 blue balls

e. there is one yellow ball

Answer 1

Solution:

4 yellow, 8 red and 3 blue balls.

Total = 4+8+3 = 15

3 balls are selected

(a):all balls are yellow

No. of ways $=^{4}C_3=4$

(b): all balls are red

No. of ways $=^{8}C_3=56$

(c): There is one ball from each color

No. of ways $=^{4}C_1\times ^{8}C_1\times ^{3}C_1=4\times 8\times 3=96$

(d): there are 2 blue balls

No. of ways $=^{3}C_2\times ^4C_1+^{3}C_2\times ^8C_1=3\times4+3\times8=36$

(e): there is one yellow ball

No. of ways $=^{4}C_1\times^8C_2+^{4}C_1\times^3C_2+^{4}C_1\times^8C_1\times^3C_1$

$=4\times 28+4\times 3+4\times 8\times 3 \\=220$