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Let XX be a discrete random variable whose probability distribution is defined as follows;

P(X=x)={k(x+1),x=1,2,3,42kx,x=5,6,70,otherwiseP(X=x)=\begin{cases}k(x+1), & x=1,2,3,4\\ 2kx, & x=5,6,7\\ 0, & \text{otherwise}\end{cases}

where kk is constant, then P(X=5)P(X=5) is equal to

Solution

✅ Correct Option: 3

Total probability must be 1:

k(2+3+4+5)+2k(5+6+7)=14k+36k=50k=1⇒k=150k(2+3+4+5)+2k(5+6+7) = 14k+36k = 50k = 1 \Rightarrow k = \frac{1}{50}.

Since x=5x=5 falls in the second branch, P(X=5)=2k(5)=10k=15P(X=5) = 2k(5) = 10k = \frac{1}{5}.

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