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Question 42 of 43

Q.The probability distribution of a random variable XX is defined as follows:
[!FORMULA] p(x)=K(x+1)!, x=1,2,3; K=Constantp(x) = \dfrac{K}{(x+1)!},\ x = 1, 2, 3;\ K = \text{Constant}
Hence find constant KK, P(3)P(3).

Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2026Subjective· 3mImportance★★★★★
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Σp(x)=K(12+16+124)=K⋅1724=1⇒K=2417\Sigma p(x)=K\left(\tfrac12+\tfrac16+\tfrac1{24}\right)=K\cdot\tfrac{17}{24}=1\Rightarrow K=\tfrac{24}{17}; P(3)=K24=117P(3)=\tfrac{K}{24}=\tfrac1{17}.

For p(x)=K(x+1)!p(x)=\dfrac{K}{(x+1)!} with x=1,2,3x=1,2,3:

p(1)=K2!=K2,p(2)=K3!=K6,p(3)=K4!=K24.p(1)=\frac{K}{2!}=\frac{K}{2},\quad p(2)=\frac{K}{3!}=\frac{K}{6},\quad p(3)=\frac{K}{4!}=\frac{K}{24}.

Since the probabilities must add to 11:

K(12+16+124)=1.K\left(\frac12+\frac16+\frac1{24}\right)=1.

Taking LCD 2424: 1224+424+124=1724\dfrac{12}{24}+\dfrac{4}{24}+\dfrac{1}{24}=\dfrac{17}{24}. So …

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