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

`(a)\ P(White) = (1)/(2)`

(b) 10 additional white balls

(c) 10 additional black balls

Explanation:

Given

`White = 5`

`Black =3`

`Red = 2`

Solving (a): P(White)

This is calculated as:

`P(White) = (White)/(Total)`

`P(White) = (5)/(5+3+2)`

`P(White) = (5)/(10)`

`P(White) = (1)/(2)`

Solving (b): Additional white balls, if
`P(White) = (3)/(4)`

Let the additional white balls be x

So:

`P(White) = (White+x)/(Total+x)`

This gives:

`(3)/(4) = (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) = (1)/(4)`

Let the additional black balls be x

So:

`P(White) = (White)/(Total+x)`

So, we have:

`(1)/(4) = (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