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1 year ago

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The newest invention of the 6.431x staff is a three-sided die. On any roll of this die, the result is 1 with probability 1/2, 2

with probability 1/4, and 3 with probability 1/4. Consider a sequence of six independent rolls of this die. Find the probability that exactly two of the rolls result in a 3.

1 answer:

Gwar [14]1 year ago

4 0

Answer:

<h2>The answer is 0.23(approx).</h2>

Step-by-step explanation:

The given die is a three sided die, hence, there are only three possibilities of getting the outcomes.

We need to find the probability of getting exactly 3s as the result.

From the sequence of 6 independent rolls, 2 rolls can be chosen in $^6C_2 = \frac{6!}{2!\times4!} = \frac{30}{2} = 15$ ways.

The probability of getting two 3 as outcome is $\frac{1}{4} \times\frac{1}{4} = \frac{1}{16}$ .

In the rest of the 4 sequences, will not be any 3 as outcome.

Probability of not getting a outcome rather than 3 is $1 - \frac{1}{4} = \frac{3}{4}$ .

Hence, the required probability is $15\times\frac{1}{16}(\frac{3}{4})^4 = \frac{1215}{4096}$ ≅0.2966 or, 0.23.

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Step-by-step explanation:

We have that:

$Area = 128$

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Such that:

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So:

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$A = (\frac{128}{y} + 2) * (y + 4)$

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Collect Like Terms

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Recall that the new dimensions are:

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So:

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To double-check;

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