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Triss [41]
2 years ago
14

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

Mathematics
1 answer:
Ierofanga [76]2 years ago
7 0
<h3>Answer: Choice B</h3>

=======================================================

Explanation:

The requirements for a probability distribution are this

  1. 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.
  2. 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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Evaluate the expression 12-3y÷2+y(2y-4÷y) for y=3
ehidna [41]

Answer:

12 - 3 y ÷ 2 + y( 2 y - 4 ÷ y )   =   21.51.

Step-by-step explanation:

Here, the given expression is:

12 - 3 y ÷ 2 + y( 2 y - 4 ÷ y )

Now, we need to evaluate the given expression for y = 3

12 - 3 y ÷ 2 + y( 2 y - 4 ÷ y )   by the rule of BODMAS

12 - 3 y ÷ 2 + y( 2 y - 4 ÷ y )  = 12 - 3(3) ÷ 2 + 3(2(3) -4 ÷ 3)  

= 12 - 9 ÷ 2 + 3( 6 - 4 ÷ 3)

=  12 - <u>9</u><u> ÷ 2</u> + 3( 6 - <u>4 ÷ </u><u>3</u>)

=  12 - <u>4.5</u> + 3( 6 - <u>1.33</u>)

= 7.5 + 3(4.67)

=  7.5 + 14.01

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