Question

A box contains eighteen $1 bills, ten $5 bills, eight $10 bills, three $20 bills, and one $100 bill. You blindly reach into the box and draw a bill at random What is the expected value of your draw? The expected value of the draw is $ (Round to the nearest cent.)

Answer 1

Answer:  $7.70

Step-by-step explanation:

Given : A box contains eighteen $1 bills, ten $5 bills, eight $10 bills, three $20 bills, and one $100 bill.

Total bills = $18+10+8+3+1=40$

We know that probability of any event =$\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

i.e. the probability of getting $1 bills =$P(E_1)=\dfrac{18}{40}=0.45$

Probability of getting $10 bills =$P(E_2)=\dfrac{10}{40}=0.25$

Probability of getting $5 bills =$P(E_3)=\dfrac{8}{40}=0.2$

Probability of getting $20 bills =$P(E_4)=\dfrac{3}{40}=0.075$

Probability of getting $100 bills =$P(E_5)=\dfrac{1}{40}=0.025$

$\text{Expected value =}1\times P(E_1)+5\times P(E_2)+10\times P(E_3)+20\times P(E_4)+100\times P(E_5)\\\\=1\times 0.45+5\times 0.25+10\times0.2+20\times 0.075+100\times 0.025\\\\=7.7$

Hence, the expected value of your draw= $7.70