When an unfair dice is thrown, the probability of getting a number $k$ on it is $\mathrm{P}(\mathrm{X}=k)=k^2 \mathrm{P}$, where $k=1,2,3,4,5,6$ and X is the random variable denoting a number on the dice, then the mean of X is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } P(X=k)=k^2 P, \sum_{i=1}^6 P(X=i)=1 \ & \Rightarrow P(X=1)+P(X=2)+\ldots .+P(X=6)=1 \ & \Rightarrow P+4 P+9 P+16 P+25 P+36 P=1 \Rightarrow P=\frac{1}{91} \ & 	ext { Mean }=1 	imes \frac{1}{91}+2 	imes \frac{4}{91}+3 	imes \frac{9}{91}+4 	imes \frac{16}{91}+5 	imes \frac{25}{91}+6 	imes \frac{36}{91} \ & \quad=\frac{441}{91} .\end{aligned}$

When an unfair dice is thrown, the probability of getting a number $k$ on it is…

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