The answer would have to be 5
Answer:
converting the equation to improper fraction
31/5+(-38/5)
31/5-38/5( the plus sign affected the minus sign inside of the bracket which is+*- = -)
multiply both terms by 5
31-38= -7
Answer:
Need pic
Step-by-step explanation:
Answer:
![E(X) =sum_{i=1}^n X_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%29%20%3Dsum_%7Bi%3D1%7D%5En%20X_i%20P%28X_i%29)
And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:
![E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%201%2A%5Cfrac%7B4%7D%7B9%7D%20-1%20%5Cfrac%7B5%7D%7B9%7D%20%3D%20-%5Cfrac%7B1%7D%7B9%7D%20%3D-0.11)
![E(X^2) =sum_{i=1}^n X^2_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3Dsum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
![E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%201%5E2%2A%5Cfrac%7B4%7D%7B9%7D%2B%20%28-1%29%5E2%20%5Cfrac%7B5%7D%7B9%7D%20%3D%201)
And the variance is defined as:
![Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%201%20-%5B-%5Cfrac%7B1%7D%7B9%7D%5D%5E2%20%3D%20%5Cfrac%7B80%7D%7B81%7D)
Step-by-step explanation:
For this case we know that we can win if we select 2 balls of the same color, so we can find the probability of win like this:
![p = \frac{Possible}{ Total}= \frac{2C1 * (5C2)}{10C2}= \frac{2*10}{45}= \frac{4}{9}](https://tex.z-dn.net/?f=%20p%20%3D%20%5Cfrac%7BPossible%7D%7B%20Total%7D%3D%20%5Cfrac%7B2C1%20%2A%20%285C2%29%7D%7B10C2%7D%3D%20%5Cfrac%7B2%2A10%7D%7B45%7D%3D%20%5Cfrac%7B4%7D%7B9%7D)
So then the probability of no win would be given by the complement:
![q = 1-p = 1- \frac{4}{9}= \frac{5}{9}](https://tex.z-dn.net/?f=%20q%20%3D%201-p%20%3D%201-%20%5Cfrac%7B4%7D%7B9%7D%3D%20%5Cfrac%7B5%7D%7B9%7D)
We can define the random variable X who represent the amount of money that we can win.
And we can use the definition of expected value given by:
![E(X) =sum_{i=1}^n X_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%29%20%3Dsum_%7Bi%3D1%7D%5En%20X_i%20P%28X_i%29)
And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:
![E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%201%2A%5Cfrac%7B4%7D%7B9%7D%20-1%20%5Cfrac%7B5%7D%7B9%7D%20%3D%20-%5Cfrac%7B1%7D%7B9%7D%20%3D-0.11)
We can calculate the second monet like this:
![E(X^2) =sum_{i=1}^n X^2_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3Dsum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
![E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%201%5E2%2A%5Cfrac%7B4%7D%7B9%7D%2B%20%28-1%29%5E2%20%5Cfrac%7B5%7D%7B9%7D%20%3D%201)
And the variance is defined as:
![Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%201%20-%5B-%5Cfrac%7B1%7D%7B9%7D%5D%5E2%20%3D%20%5Cfrac%7B80%7D%7B81%7D)