Answer:
[tex](a)\ P(White) = \frac{1}{2}[/tex]
(b) 10 additional white balls
(c) 10 additional black balls
Step-by-step explanation:
Given
[tex]White = 5[/tex]
[tex]Black =3[/tex]
[tex]Red = 2[/tex]
Solving (a): P(White)
This is calculated as:
[tex]P(White) = \frac{White}{Total}[/tex]
[tex]P(White) = \frac{5}{5+3+2}[/tex]
[tex]P(White) = \frac{5}{10}[/tex]
[tex]P(White) = \frac{1}{2}[/tex]
Solving (b): Additional white balls, if [tex]P(White) = \frac{3}{4}[/tex]
Let the additional white balls be x
So:
[tex]P(White) = \frac{White+x}{Total+x}[/tex]
This gives:
[tex]\frac{3}{4} = \frac{5+x}{10+x}[/tex]
Cross multiply
[tex]30+3x = 20 + 4x[/tex]
Collect like terms
[tex]4x - 3x = 30 - 20[/tex]
[tex]x = 10[/tex]
Hence, 10 additional white balls must be added
Solving (c): Additional black balls, if [tex]P(White) = \frac{1}{4}[/tex]
Let the additional black balls be x
So:
[tex]P(White) = \frac{White}{Total+x}[/tex]
So, we have:
[tex]\frac{1}{4} = \frac{5}{10+x}[/tex]
Cross multiply
[tex]10+x = 5 * 4[/tex]
[tex]10+x = 20[/tex]
Collect like terms
[tex]x = 20 -10[/tex]
[tex]x = 10[/tex]
Hence, 10 additional black balls must be added
