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Q.A random variable X has the following probability distribution: XX takes values 0,1,2,3,4,5,6,70, 1, 2, 3, 4, 5, 6, 7 with P(X)=0, k, 2k, 2k, 3k, k2, 2k2, 7k2+kP(X) = 0,\ k,\ 2k,\ 2k,\ 3k,\ k^2,\ 2k^2,\ 7k^2 + k respectively. Then the value of kk is ___.

(a) −1-1
(b) 11
(c) 110\frac{1}{10}
(d) −110-\frac{1}{10}
Gujarat GsebGSEB Higher Secondary Certificate (HSC) Examination 2022MCQ· 1mImportance★★★★★
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All probabilities must sum to 1; solve the resulting quadratic and keep the valid root.

∑P(X)=0+k+2k+2k+3k+k2+2k2+(7k2+k)=10k2+9k=1.\sum P(X) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2+k) = 10k^2 + 9k = 1.

10k2+9k−1=0⇒k=−9±81+4020=−9±1120.10k^2 + 9k - 1 = 0 \Rightarrow k = \dfrac{-9\pm\sqrt{81+40}}{20} = \dfrac{-9\pm11}{20}.

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