AP EAMCET · Maths · Probability
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
- A 25
- B 5
- C \(\frac{441}{9}\)
- D \(\frac{441}{91}\)
Answer & Solution
Correct Answer
(D) \(\frac{441}{91}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } 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} \\ & \text { Mean }=1 \times \frac{1}{91}+2 \times \frac{4}{91}+3 \times \frac{9}{91}+4 \times…
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