Aspire Faculty ID #16203 · Topic: NIMCET 2011 · Just now
NIMCET 2011

A random variable $X$ has the probability distribution: \[\begin{array}{c|ccccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X=x) & a & 3a & 5a & 7a & 9a & 11a & 13a & 15a & 17a \end{array} \]The value of $a$ is:

Solution

$ \text{Total probability} = 1 $ $ a(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) = 1 $ $ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81 $ $ 81a = 1 $ $ a = \frac{1}{81} $

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