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  • Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 4 sub question 1 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 4 sub question 1 maths textbook solution

Answers (1)

Answer: So, the possible value of c=\frac{1}{3}

Hint: \sum P\left ( x=n \right )=1

Given:

X 0 1 2
P\left ( X \right ) 3c^{3} 4c-10c^{2} 5c-1

where c> 0

Solution : 
Since \sum P\left ( x \right )=1

\Rightarrow P(0)+P(1)+P(2)=1

\Rightarrow 3 c^{3}+4 c-10 c^{2}+5 c-1=1

\Rightarrow 3 c^{3}-10 c^{2}+9 c-2=0

\Rightarrow 3 c^{3}-3 c^{2}-7 c^{2}+7 c+2 c-2=0

\Rightarrow 3 c^{2}(c-1)-7 c(c-1)+2(c-1)=0

\Rightarrow(c-1)\left(3 c^{2}-7 c+2\right)=0

\Rightarrow(c-1)\left(3 c^{2}-6 c-c+2\right)=0

\Rightarrow(c-1)(3 c-1)(c-2)=0

c=1, c=2, \quad c=\frac{1}{3}

Only c=\frac{1}{3} is possible, because if c=1  or  c=2 then  P\left ( 1 \right )  will become negative.

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