# How can you desing a dice where the probability of 2 occurs is twice the probability of 1 occurs, the probability of 3 occurs is three times the probability of 1 occurs, the probability of 4 occurs is four times the probability of 1 occurs, the probability of 5 occurs is five times the probability of 1 occurs, and the probability of 6 occurs is six times the probability of 1 occurs?

How can you desing a dice where the probability of 2 occurs is twice the probability of 1 occurs, the probability of 3 occurs is three times the probability of 1 occurs, the probability of 4 occurs is four times the probability of 1 occurs, the probability of 5 occurs is five times the probability of 1 occurs, and the probability of 6 occurs is six times the probability of 1 occurs?
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Step 1
Let probability of 1 occurs = x
$⇒P\left(1\right)=x$
Then from given conditions
$P\left(2\right)=2x\phantom{\rule{0ex}{0ex}}P\left(3\right)=3x\phantom{\rule{0ex}{0ex}}P\left(4\right)=4x\phantom{\rule{0ex}{0ex}}P\left(5\right)=5x\phantom{\rule{0ex}{0ex}}P\left(6\right)=6x\phantom{\rule{0ex}{0ex}}\because P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)=1\phantom{\rule{0ex}{0ex}}⇒x+2x+3x+4x+5x+6x=1\phantom{\rule{0ex}{0ex}}21x=1\phantom{\rule{0ex}{0ex}}x=\frac{1}{21}\phantom{\rule{0ex}{0ex}}\therefore P\left(1\right)=\frac{1}{21}\phantom{\rule{0ex}{0ex}}P\left(2\right)=\frac{2}{21}\phantom{\rule{0ex}{0ex}}P\left(3\right)=\frac{3}{21}\phantom{\rule{0ex}{0ex}}P\left(4\right)=\frac{4}{21}\phantom{\rule{0ex}{0ex}}P\left(5\right)=\frac{5}{21}\phantom{\rule{0ex}{0ex}}P\left(6\right)=\frac{6}{21}$