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The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is

\[ \begin{array} (A) 1 \quad & (B) 2 \quad & (C) 5 \quad &(D) \frac{8}{3} \end{array} \]

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  • Mean of the probability distribution = $\sum (X_i \times P(X_i))$ The Expected value of X is nothing but the mean of X.
In a die toss, the total number of outcomes = 6. Let X be the random variable representing a number on the die.
The # 1 appears on 3 faces, P (X=1) = $\large\frac{3}{6}$
The # 2 appears on 2 faces, P (X=2) = $\large\frac{2}{6}$
The # 5 appears on 1 face, P (X=5) = $\large\frac{1}{6}$
Given this distribution, we can calculate the Mean =\(\;1\times\large\frac{3}{6}\)\(+2\times\large\frac{2}{6}\)\(+5\times\large\frac{1}{6}\) = \(\large\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\) $\;=\large\frac{12}{6} $$=2$
answered Jun 20, 2013 by balaji.thirumalai

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