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Please help me I have been stuck on this question for ages

Please help me I have been stuck on this question for ages-example-1

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Answer:28 green and 35 red

Explanation:

Given

If there are r red counter and g green counter then

Probability of drawing a green counter is
P(g)=(4)/(9)

and
P(g)=\frac{\text{No of g counter}}{\text{Total no of counter}}

Thus
\frac{\text{No of g counter}}{\text{Total no of counter}}=(4)/(9)


(g)/(g+r)=(4)/(9)


\Rightarrow 9g=4g+4r


\Rightarrow 5g=4r\quad \ldots(i)

Also if 4 red and 2 green counter is added the probability of drawing a green counter is


P(g)=(10)/(23)=\frac{\text{No of g counter}}{\text{Total no of counter}}


\Rightarrow (10)/(23)=(g+2)/(g+2+r+4)


\Rightarrow (10)/(23)=(g+2)/(g+r+6)


\Rightarrow 10g+10r+60=23g+46


\Rightarrow 10r+14=13g\ quad \ldots(ii)

Substitute the value of g in equation (ii)[/tex]


\Rightarrow 10* (5)/(4)g+14=13g


\Rightarrow (25)/(2)g+14=13g


\Rightarrow g=28

Therefore
r=35

Thus there 28 green counter and 35 red counter

User Eric Andres
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