# Math - Help!

The sample space for a roll of two number cubes is shown in the table.

(1,1)|(1,2)|(1,3)|(1,4)|(1,5),(1,6)
(2,1)|(2,2)|(2,3)|(2,4)|(2,5)|(2,6)
(3,1)|(3,2)|(3,3)|(3,4)|(3,5)|(3,6)
(4,1)|(4,2)|(4,3)|(4,4)|(4,5)|(4,6)
(5,1)|(5,2)|(5,3)|(5,4)|(5,5)|(5,6)
(6,1)|(6,2)|(6,3)|(6,5)|(6,5)|(6,6) What is the probability that the roll will result in one even and one odd number?

A. 1/9
B. 1/4
C. 1/3
D. 1/2 ***

The two numbers rolled can be added to get a sum. Find P(sum is greater than 5).

A. 5/6 ***
B. 13/18
C. 2/3
D. 1/2

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1. A quality supervisor for a factory that makes cell phone cases examines a sample of 46 cases. She finds that 8 cases are the wrong size. Which is the best prediction of the number of cases that will be the wrong size in a shipment of 3,000?

A. 410
B. 475
C. 506
D. 522 ***

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3. first one is correct

prob(sum > 5)
= 1 - prob(sum ≤ 5)
= 1 - 10/36
= 26/36
= 13/18

One might think it has been answered and it gets overlooked ....

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4. Bobpursley said I was wrong on the second one, I ORIGINALLY got 13/18. And, he said I got it wrong. But, I did. Click on my name, and go to "Are all answers right?" Strange, but thank you.

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5. It said: P(sum is greater than 5)
so we have to exclude those equal or less than 5

11 12 13 14
21 22 23
31 32
41 -----> there are 10 of those, leaving 26 to be > 5

so prob(sum> 5)
= 1 - 10/36 = 26/36 = 13/18
or by using the 26 count directly
prob( sum > 5) = 26/36 = 13/18

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6. I trust you, too. I, too, got the same answer. But, bobpursley unexpectedly thought it was wrong or something.

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7. Sandy, I have made many errors and mistakes.
Sometimes we read a question too quickly and miss something.
In my case I just call it a "senior moment, lol

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