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
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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) = 3636
The # 2 appears on 2 faces, P (X=2) = 2626
The # 5 appears on 1 face, P (X=5) = 1616
Given this distribution, we can calculate the Mean =\(\;1\times\large\frac{3}{6}\)\(+2\times\large\frac{2}{6}\)\(+5\times\large\frac{1}{6}\) = \(\large\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\) =126=126=2
Mean of the probability distribution = ∑(Xi×P(Xi))∑(Xi×P(Xi)) The Expected value of X is nothing but the mean of X.
The # 1 appears on 3 faces, P (X=1) = 3636
The # 2 appears on 2 faces, P (X=2) = 2626
The # 5 appears on 1 face, P (X=5) = 1616
Given this distribution, we can calculate the Mean =\(\;1\times\large\frac{3}{6}\)\(+2\times\large\frac{2}{6}\)\(+5\times\large\frac{1}{6}\) = \(\large\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\) =126=126=2
Mean of the probability distribution = ∑(Xi×P(Xi))∑(Xi×P(Xi)) The Expected value of X is nothing but the mean of X.
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