Which of the following could be the probability distribution for a loaded number cube?

Mathematics

1 answer:

Ierofanga [76]3 years ago

7 0

<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 $0 \le p \le 1$ 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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$\displaystyle \sum_{n=11}^{20} a_n = 300 \\ (a-d) \sum_{n=11}^{20} 1 + d \sum_{n=11}^{20} n = 300 \\ (a-d) \left(\sum_{n=1}^{20} 1 - \sum_{n=1}^{10} 1\right) 1 + d \left(\sum_{n=1}^{20} n - \sum_{n=1}^{10} n\right) = 300 \\ 10a+145d = 300$