Suppose we have a bag with 10 slips of paper in it. Eight of these have a 2 on them and the other two have a 4 on them.
What is the expected value of the number shown when we draw a single slip of paper?
I appreciate the help but your answer was wrong. here is the answer
Expected value is computed by multiplying all of the possible outcomes by their probabilities, and then summing. \(8/10\) of the time we draw a 2 and \(2/10\) of the time we draw a 4. Therefore the expected value is
\(2\cdot\frac{8}{10} + 4\cdot\frac{2}{10} = \frac{12}{5}.\)
\(P[2]=\dfrac{8}{10}=\dfrac{4}{5}\\ P[4] = \dfrac 1 5\\ E[N] = 8\cdot 2 \cdot \dfrac 4 5 + 2 \cdot 4 \cdot \dfrac 1 5 = \\ \dfrac{64}{5} + \dfrac 8 5 = \dfrac{72}{5}\)
.I appreciate the help but your answer was wrong. here is the answer
Expected value is computed by multiplying all of the possible outcomes by their probabilities, and then summing. \(8/10\) of the time we draw a 2 and \(2/10\) of the time we draw a 4. Therefore the expected value is
\(2\cdot\frac{8}{10} + 4\cdot\frac{2}{10} = \frac{12}{5}.\)