Which of the following could be the probability distribution for a loaded number cube?
1 answer:
<h3>Answer: Choice B</h3>
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Explanation:
The requirements for a probability distribution are this
- The individual probabilities must be between 0 and 1, inclusive of both endpoints. We can say
where p is an individual probability.
- The probabilities must sum to 1.
Condition #1 shown above allows us to rule out choice C due to -0.10 not being in the interval from 0 to 1.
We can also rule out choice D since the probabilities sum to this
0.20+0.15+0.20+0.20+0.20+0.20 = 1.15
The sum is too large. The sum needs to be 1.00 or just 1.
So that leaves choice A or choice B as the possible answer. Both distributions fit conditions #1 and #2 as shown above (I'll let you confirm these facts). However, choice A is ruled out because that distribution is considered fair. Each probability for choice A is equal, so each side of the cube is equally likely to be landed on.
For choice B, the probabilities are different, so we don't have a fair cube here and it is considered loaded or biased.
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Rx - sx + y = b
WHEN SOLVING FOR X :
rx - sx + y = b
We must get x onto it's own side, so subtract y from both side.s
rx - sx = b - y
Then, factor out x.
x(r - s) = b - y
Then, divide both sides by (r - s).
x(r - s) ÷ (r - s) = b - y ÷ (r - s)
Simplify.
x = b - y / r - s →
WHEN SOLVING FOR Y :
rx - sx + y = b
We need to isolate y, so get rid of everything BUT y on the left side.
Subtract rx from both sides.
-sx + y = b - rx
Then, add sx to both sides.
y = b - rx + sx
~Hope I helped!~
WHEN SOLVING FOR X :
rx - sx + y = b
We must get x onto it's own side, so subtract y from both side.s
rx - sx = b - y
Then, factor out x.
x(r - s) = b - y
Then, divide both sides by (r - s).
x(r - s) ÷ (r - s) = b - y ÷ (r - s)
Simplify.
x = b - y / r - s →
WHEN SOLVING FOR Y :
rx - sx + y = b
We need to isolate y, so get rid of everything BUT y on the left side.
Subtract rx from both sides.
-sx + y = b - rx
Then, add sx to both sides.
y = b - rx + sx
~Hope I helped!~