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sasho [114]
2 years ago
14

You roll two fair dice. If they land with a sum of 7, you get a $10. If they land with a sum of 6 or 8, you get $5. Everything e

lse earns $0. Which of the following would be the highest value you could set the price and the player would still win?
$8
$5
$2
$10​
Mathematics
1 answer:
Sindrei [870]2 years ago
3 0

Using the expected value, it is found that the highest value you could set the price is: $5.

-----------------------------

  • A game is fair, that is, the player would still win, if the <u>expected value is 0</u>.
  • The expected value is the <u>sum of each outcome multiplied by it's probability</u>.
  • A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.

  • For two fair dice, there are 36 total outcomes, as 6^2 = 36.
  • The charge is x.
  • 6 outcomes result in a sum of 7, which are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), thus, \frac{6}{36} probability of getting 10.
  • 10 outcomes result in a sum of 6 or 8, which are (1,5), (2,4), (2,6), (3,3), (3,5), (4,2), (4,4), (5,1), (5,3) and (6,2), thus, \frac{8}{36} probability of getting x.
  • 36 - 14 = 22, thus \frac{22}{36} probability of losing x.

Since we want the expected value to be 0:

\frac{6}{36}(10) + \frac{8}{36}(5) - \frac{22}{36}x = 0

\frac{60 + 50 - 22x}{36} = 0

22x = 110

x = \frac{110}{22}

x = 5

The highest value you could set the price is: $5.

A similar problem is given at brainly.com/question/24905256

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Answer:

<h2><u><em>6. </em></u></h2><h2><u><em>   a. 11 and 2/3 yds. squared</em></u></h2><h2 /><h2><u><em>   b. Yes, the volume of the shed is 11 and 1/3 yards squared and what she's trying to put into it is only 10 yards squared, if put in properly, it will be able to fit.</em></u></h2><h2 /><h2><u><em>7. 1,110 in. squared</em></u></h2>

Step-by-step explanation:

6.

a.

(10/3)*(14/9)*(9/4)

= 11 2/3

b.

Yes, because the volume of the shed is about 11.67 yards long, the 10 yards of wood will fit in the shed.

7.

For this one, we have to break it into two pieces.

(I made them into a small box and and big box)

The measurements of the small box are 7*5*6.

The measurements of the big box are 20*5*9.

Using this information, we can make the following equation and solve it quickly.

(7*5*6) + (20*5*9)

(210) + (900)

1,110

Thus, the volume of this box is 1,110 in. ^2

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