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Answer: Choice D) -$22</h3>
You'll lose on average $22 per roll.
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Explanation:
Normally there is a 1/6 chance to land on any given side of a standard die, but your friend has loaded the die in a way to make it have a 40% chance to land on "1" and an equal chance to land on anything else. Since there's a 40% chance to land on "1", this leaves 100% - 40% = 60% for everything else.
Let's define two events
- A = event of landing on "1".
- B = event of landing on anything else (2 through 6).
So far we know that P(A) = 0.40 and P(B) = 0.60; I'm using the decimal form of each percentage.
The net value of event A, which I'll denote as V(A), is -100 since you pay $100 when event A occurs. So we'll write V(A) = -100. Also, we know that V(B) = 30 and this value is positive because you receive $30 if event B occurs.
To recap things so far, we have the following:
- P(A) = 0.40
- P(B) = 0.60
- V(A) = -100
- V(B) = 30
Multiply the corresponding probability and net value items together
- P(A)*V(A) = 0.40*(-100) = -40
- P(B)*V(B) = 0.60*30 = 18
Then add up those products:
-40+18 = -22
This is the expected value, and it represents the average amount of money you earn for each dice roll. So you'll lose on average about $22. Because the expected value is not zero, this means this game is not mathematically fair.
This does not mean that any single die roll you would lose $22; instead it means that if you played the game say 1000 or 10,000 times, then averaging out the wins and losses will get you close to a loss of $22.
Answer:
Imagine playing R0bl0x
Step-by-step explanation:
( dont take it srsly lol im jk)
Answer:
Step-by-step explanation:
A=πr2
Circumference = 2πr
20π = 2πr
r=10
π=3.14
A=3.14 * 10^2
=314 cm^2
