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Lorico [155]
3 years ago
11

3. A magical four-sided die is rolled twice. Let S be the sum of the resultsof the two rolls. We are told that the probability t

hat S = k is proportional to k,for k = 2, 3, . . . , 8, and that all possible ways that a given sum k can arise are equallylikely. Construct an appropriate probabilistic model and find the probability of gettingdoubles.
Mathematics
1 answer:
Fed [463]3 years ago
6 0

Answer:

P(S=k) = (1/35)k

8/21

Step-by-step explanation:

P(S=k) is proportional to k

P(S=k) = ck

Total probability is 1

k = 2,3.... 8

Sum of all probabilities =

2c + 3c + 4c + 5c + 6c + 7c + 8c = 1

35c = 1

c = 1/35

P(S=k) = (1/35)k, where k = 2,3,...8

P(1&1):

P(S=2) = 2/35

P(2&2):

P(S=4)/3 = (1/3)(4/35) = 4/105

Because for S=4: (1,3) (2,2) (3,1) and all are equally likely

P(3&3):

P(S=6)/3 = (1/3)(6/35) = 2/35

Because for S=6: (2,4) (3,3) (4,2)

P(4&4):

P(S=8) = 8/35

P(getting doubles = 2/35 + 4/105 + 2/35 + 8/35 = 40/105 = 8/21

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lys-0071 [83]
So if we want to know the common solution(s) to a system of 2 equations, So we can just set both equations equal to each other and solve for the x value(s). That’s where I start below;

2x^2-13x+21 = 2x^2+9x-56
2x^2 cancels out and moving everything to one side and anything with an x variable to the other side we have then;
-22x=-77
22x=77 by cancelling the negative signs
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Hope this helps you. Any questions please ask.
4 0
3 years ago
(10-3)*2+(5-14/2 help
Dafna11 [192]
2(10-3) distribute = 2*10 is 20 and 2*3 is 6
20-6+(5-14/2)
now 20-6 is 14
14+(5-14/2)
now 14/2 is 7
14+(5-7)
now 5-7 is -2
14+-2
is 12 
the answer is 12
but if you want to you can do 12/3 to get the greatest common factor 
which is 4
hope i help you :P 
6 0
3 years ago
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Oxnard Petro Ltd. is buying hurricane insurance for its off-coast oil drilling platform. During the next five years, the probabi
Andrew [12]

Answer:

The expected loss is $275 million.

Step-by-step explanation:

Expected loss can be determined as the sum of the product of each possible loss by the its probability of occurence. In this situation, there are only two possible losses listed since the probability of no loss doesn't add any value to the expected loss and should be disregarded.

Expected loss (in millions) = EL

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The expected loss is $275 million.

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Explanation is in the file

tinyurl.com/wpazsebu

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