Question

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?

Answer 1

Answer:

$(a)\ P(White) = \frac{1}{2}$

(b) 10 additional white balls

(c) 10 additional black balls

Step-by-step explanation:

Given

$White = 5$

$Black =3$

$Red = 2$

Solving (a): P(White)

This is calculated as:

$P(White) = \frac{White}{Total}$

$P(White) = \frac{5}{5+3+2}$

$P(White) = \frac{5}{10}$

$P(White) = \frac{1}{2}$

Solving (b): Additional white balls, if $P(White) = \frac{3}{4}$

Let the additional white balls be x

So:

$P(White) = \frac{White+x}{Total+x}$

This gives:

$\frac{3}{4} = \frac{5+x}{10+x}$

Cross multiply

$30+3x = 20 + 4x$

Collect like terms

$4x - 3x = 30 - 20$

$x = 10$

Hence, 10 additional white balls must be added

Solving (c): Additional black balls, if $P(White) = \frac{1}{4}$

Let the additional black balls be x

So:

$P(White) = \frac{White}{Total+x}$

So, we have:

$\frac{1}{4} = \frac{5}{10+x}$

Cross multiply

$10+x = 5 * 4$

$10+x = 20$

Collect like terms

$x = 20 -10$

$x = 10$

Hence, 10 additional black balls must be added