Question

400 draws are made at random with replacement from 5 tickets that are marked -2, -1, 0, 1, and 2 respectively. Find the expected value of the sum of the positive numbers drawn

Answer 1

The probability of drawing a ticket marked 1 on each draw is 1/5.
The probability of drawing a ticket marked 2 on each draw is 1/5.
The expected number of tickets marked 1 drawn in 400 draws is\[400\times \frac{1}{5}\]
The expected number of tickets marked 2 drawn in 400 draws is
\[400\times \frac{1}{5}\]
The expected value of the sum of the positive numbers drawn is
\[(1\times 400\times \frac{1}{5})+(2\times 400\times \frac{1}{5})=you\ can\ calculate\]

Answer 2

240

Answer 3

ok i have now seen where i was wrong. i just obtained the number of times a positive appears which=np=(2/5)×400=160. n later multiplied by the average of the positives. ie (1+2)/2. Thanks for showing me the way !!!

Answer 4

You're welcome :)

Answer 5

So The expected value of the sum of the all drawn is ? \[ \left( 1*400*\frac{ 1 }{ 5 } \right)+\left( 2*400*\frac{ 1 }{ 5 } \right)+\left( 3*400*\frac{ 1 }{ 5 } \right)+\left( 4*400*\frac{ 1 }{ 5 } \right)+\left( 5*400*\frac{ 1 }{ 5 } \right)\]

Answer 6

all the numbers drawn