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Scrat [10]
3 years ago
8

A pair of fair dice will be rolled. Let the random variable X = the sum of the numbers shown on their faces. The probability dis

tribution of X is given in the table.
A 2-column table with 11 rows. Column 1 is labeled sum with entries 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Column 2 is labeled probability with entries 0.028, 0.056, 0.083, 0.111, 0.139, 0.167, 0.139, 0.111, 0.083, 0.056, 0.028.
What is the probability that the dice show a sum of at least 6?

P(X ≥
) =
Mathematics
1 answer:
Afina-wow [57]3 years ago
7 0

Answer: 12 and 0.972

Step-by-step explanation: am doing it

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Z=1/218(x-810) solve this please
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-3 78⁄109

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The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris a
nadezda [96]

Answer and explanation:

Given : The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0, 0.3, 0.4, and 0.6, respectively.

The probabilities of no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0.8, 0.02, 0.08, and 0.1, respectively.

Let the event E denote the poor print quality.

Let the event A be the no printer problem i.e. P(A)=0.8

Let the event B be the misaligned paper i.e. P(B)=0.02

Let the event C be the high ink viscosity i.e. P(C)=0.08

Let the event D be the printer-head debris i.e. P(D)=0.1

and the probabilities of poor print quality given printers are

P(E|A)=0,\ P(E|B)=0.3,\ P(E|C)=0.4,\ P(E|D)=0.6

First we calculate the probability that print quality is poor,

P(E)=P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)+P(D)P(E|D)

P(E)=(0)(0.8)+(0.3)(0.02)+(0.4)(0.08)+(0.6)(0.1)

P(E)=0+0.006+0.032+0.06

P(E)=0.098

a. Determine the probability of high ink viscosity given poor print quality.

P(C|E)=\frac{P(E|C)P(C)}{P(E)}

P(C|E)=\frac{0.4\times 0.08}{0.098}

P(C|E)=\frac{0.032}{0.098}

P(C|E)=0.3265

b. Given poor print quality, what problem is most likely?

Probability of no printer problem given poor quality is

P(A|E)=\frac{P(E|A)P(A)}{P(E)}

P(A|E)=\frac{0\times 0.8}{0.098}

P(A|E)=\frac{0}{0.098}

P(A|E)=0

Probability of misaligned paper given poor quality is

P(B|E)=\frac{P(E|B)P(B)}{P(E)}

P(B|E)=\frac{0.3\times 0.02}{0.098}

P(B|E)=\frac{0.006}{0.098}

P(B|E)=0.0612

Probability of printer-head debris given poor quality is

P(D|E)=\frac{P(E|D)P(D)}{P(E)}

P(D|E)=\frac{0.6\times 0.1}{0.098}

P(D|E)=\frac{0.06}{0.098}

P(D|E)=0.6122

From the above conditional probabilities,

The printer-head debris problem is most likely given that print quality is poor.

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Answer:

A

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