Expected value of drawing bills

expected valueprobabilityrandom variables

There are 2 $\$10$ bills, 2 $\$20$ bills, and 1 $\$50$ bill inside a box. You draw four bills at random. What is the expected value of amount that you have (assuming draws are without replacement)?

My approach: Expected value of bills from the box is $2/5*10 + 2/5*20 + 1/5*50 = 22$
I have a couple of questions at this point-

  1. If draws are with replacement, then the expected value is simply 22*4 by linearity of expectations, am I correct?
  2. How to approach this when the draws are without replacement?

Edit: I have corrected the arithmetic error as pointed out in the comments and the answer.

Best Answer

  1. If draws are with replacement, then the expected value is simply 18*4 by linearity of expectations, am I correct?

The principle of multiplying the expected value for one draw by four is correct. That's a nice observation of the independence between the draws. However, be careful with the arithmetic.

Let $\$X$ be the value of the bill drawn.

$$\text{expected value of the bill} = \$E[X] = \$10 \cdot P(X = 10) + \$20 \cdot P(X = 20) + \$50 \cdot P(X = 50) \\ = \$\left(10 \cdot \frac25+ 20 \cdot \frac25 + 50 \cdot \frac15\right) = \$(4 + 8 + 10) = \$22$$

Hence the expected value of drawing four bills with replacement is $\$4E[X] = \$4(22) = \$88$.

  1. How to approach this when the draws are without replacement?

You have five bills in total. When you draw four bills from them, only one bill is left, so the answer is simple: $\$(10\cdot2+20\cdot2+50\cdot1-E[X]) = \$(110 - 22) = \$88$.

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