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Question 36 of 41

Q.The probability distribution function of a discrete random Variable XX is : f(x)={2k,x=13k,x=34k,x=50,otherwisef(x)=\begin{cases}2k, & x=1\\3k, & x=3\\4k, & x=5\\0, & \text{otherwise}\end{cases} where kk is some constant, find kk.

Puducherry TnboardTamil Nadu HSC (DGE) Commerce Board 2025Subjective· 2mImportance★★★★★
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A probability mass function must satisfy ∑f(x)=1\sum f(x)=1; here 2k+3k+4k=9k=12k+3k+4k=9k=1, giving k=19k=\tfrac19.

Apply the total-probability condition. For a discrete random variable, ∑xf(x)=1\sum_x f(x)=1:

f(1)+f(3)+f(5)=2k+3k+4k=1.f(1)+f(3)+f(5)=2k+3k+4k=1.

Solve.

9k=1  ⟹  k=19.9k=1\implies k=\frac{1}{9}.

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