+0

0
42
4

There are three $20 bills, six$10 bills, seven $5 bills, and four$1 bills in your wallet. You select one bill at random. What is the expected value for this experiment?

$_______ You roll a pair of six-sided number cubes, numbered 1 through 6. What is the probability of rolling two odd numbers? (Enter your probability as a fraction.) =__________ Guest Nov 9, 2018 #1 +13718 +2 Twenty bills 3/20 ($20)  + 6/20 (10) + 7/20(5) + 4/20(1) = 7.95 expected value

ElectricPavlov  Nov 9, 2018
#2
+2793
+2

$$\text{There are 20 total bills. Assume we choose any of them with equal probability.}\\ P[20] = \dfrac{3}{20}\\ P[10]= \dfrac{6}{20}=\dfrac{3}{10}\\ P[5] = \dfrac{7}{20}\\ P[1] = \dfrac{4}{20} = \dfrac{1}{5} \\ \text{Let }X \text{ be the value of the bill chosen.}\\ E[X] = 20\cdot \dfrac{3}{20} + 10\cdot \dfrac{3}{10} + 5 \cdot \dfrac{7}{20} + 1\cdot \dfrac{1}{5}\\ E[X] = \dfrac{159}{20} = 7.95$$

Rom  Nov 9, 2018
edited by Rom  Nov 9, 2018
#3
+13718
+2

3/6 are odd for each die

3/6 x 3/6 = 9/36 = 1/4 chance of both being odd

ElectricPavlov  Nov 9, 2018
#4
+2793
0

deleted

Rom  Nov 9, 2018
edited by Rom  Nov 9, 2018