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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?

User JoshMB
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1 Answer

3 votes

Answer:


(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

User Abiola
by
8.3k points

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