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An examination consists of 4 papers. 1st 3 papers has $n$ marks and 4th paper has $2n$ marks what is the probability that a student obtain $3n$ marks in 4 papers.

ROBINSON
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1 Answers1

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This is equivalent to solving: $$a+b+c+d=3n$$ where$$a,b,c=[0,n]$$ $$d=[0,2n]$$ and then use this :Link

rnjai
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