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A box contains 5 white balls, 3 black balls, and 2 red balls.A-What is the probability of drawing a white ball?B- How many white balls must be added to the box so that the probability of drawing a white ball is 3/4?C-How many black balls must be added to the original assortment so that the probability of drawing a white ball is 1/4?

Sagot :

MrRoyal

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

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