Question

In an urn, there are 30 blue balls, 20 red balls and x green balls and y black balls. A ball is taken at random from the urn. The probability that the chosen ball is not red in colour is $\frac{3}{4}$ and it is known that the number of black balls is twice the number of green balls. Find the value of y-x.

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Answer 1

Number of blue balls = 30

Number of red balls = 20

Number of green balls = x

Number of black balls = y

Total number of balls = 30+ 20 + x + y = 50 + x + y

$\frac{1}{4}=\frac{20}{50+x+y}$

Subtract $\frac{3}{4}$ from both sides and rearrange,

$\frac{1}{4}=\frac{20}{50+x+y}$

Cross multiplying

50 + x + y = 80

Or, x + y = 30 ...........(i)

It is given that the number of black balls (y) is twice as many as green balls (x)

i.e y = 2x ...........(ii)

Substituting (ii) in Equation (i)
we have x + 2x = 30
3x = 30
x = 10
From (ii)

y = 2( 10) = 20
$\therefore y–x=20-10=10$