Answer:
b) Binomial
c) Poisson
Step-by-step explanation:
The geometric distribution is the number of trials required to have r successes. The measures the number of sucesses(wins), not the number of trials required to win r games. So the geometric distribution does not apply.
For each match, there are only two possible outcomes, either the skilled player wins, or he does not. The probability of the skilled player winning a game is independent of other games. So the binomial distribution applies.
We can also find the expected number of wins of the skilled player, which is 15*0.9 = 13.5. The Poisson distribution is a discrete distribution in which the only parameter is the expected number of sucesses. So the Poisson distribution applies.
So the correct answer is:
b) Binomial
c) Poisson
<h3>
Answer: Choice D) -$22</h3>
You'll lose on average $22 per roll.
====================================================
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.
I'm pretty sure that this is a trick question, the answer is 61%.
Step-by-step explanation:
Sorry to say, but there is no picture...
It is true that the product of two consecutive even integers are always one less than the square of their average.
<u>Step-by-step explanation</u>:
Let the two consecutive odd integers be 1 and 3.
- The product of 1 and 3 is (1
3)=3 - The average of 1 and 3 is (1+3)/2 =4/2 = 2
- The square of their average is (2)² = 4
∴ The product 3 is one less than the square of their average 4.
Let the two consecutive even integers be 2 and 4.
- The product of 2 and 4 is (24)=8
- The average of 2 and 4 is (2+4)/2 =6/2 = 3
- The square of their average is (3)² = 9
∴ The product 8 is one less than the square of their average 9.
Thus, It is true that the product of two consecutive even integers are always one less than the square of their average.
