On the first roll, you have a 1/3 probability of rolling a 1 or 2 and thus winning $200.
There's a 2/3 probability of not rolling a 1 or 2 on the first roll. On the second roll, there is again a 1/3 probability of rolling a 1 or 2, and 2/3 probability otherwise, so that there is a 1/3*2/3 = 2/9 probability of getting a 1 or 2 and thus winning $100, and 2/3*2/3 = 4/9 probability of losing.
(a) Let
be a random variable representing the winnings from playing the game. It has the probability mass function
![P(W=w)=\begin{cases}\frac13&\text{for }w=\$200\\\frac29&\text{for }w=\$100\\\frac49&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=P%28W%3Dw%29%3D%5Cbegin%7Bcases%7D%5Cfrac13%26%5Ctext%7Bfor%20%7Dw%3D%5C%24200%5C%5C%5Cfrac29%26%5Ctext%7Bfor%20%7Dw%3D%5C%24100%5C%5C%5Cfrac49%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
(b) Compute the expected value of
:
![E[W]=\displaystyle\sum_w w\,P(W=w)=\$200\cdot\frac13+\$100\cdot\frac29+\$0\cdot\frac49\approx\$88.89](https://tex.z-dn.net/?f=E%5BW%5D%3D%5Cdisplaystyle%5Csum_w%20w%5C%2CP%28W%3Dw%29%3D%5C%24200%5Ccdot%5Cfrac13%2B%5C%24100%5Ccdot%5Cfrac29%2B%5C%240%5Ccdot%5Cfrac49%5Capprox%5C%2488.89)
Answer:
the tird one
Step-by-step explanation:
the asking you like for the number who repeat I hope it help if need more explanation tell me
Using mixture formula:
Let volume be v, % be p
1 should be the first mixture, 2 should be the second mixture, 3 should be the final mixture.
p₁v₁ + p₂v₂ = p₃v₃
p₁ = 25%, v₁ = 40 p₂ = 60%, v₂ = ? p₃ = 30%, v₃ =?
Let v₂ = x, v₃ = total volumes = v₁ + v₂ = (40 + x)
Using:
p₁v₁ + p₂v₂ = p₃v₃
25*40 + 60*x = 30*(40 + x)
1000 + 60x = 1200 + 30x
60x - 30x = 1200 - 1000
30x = 200
x = 200/30
x = 20/3
Recall x = v₂, so she has to add 20/3 milliters of the 60% solution in order to obtain 30% solution.
I hope this explains it.
Answer:
-7
Step-by-step explanation: