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Orlov [11]
3 years ago
6

Calculate the expected value E(X) of the given random variable X. X is the higher number when two dice are rolled, or the common

number if doubles are rolled. (So, a roll of 4-3 would be given a value of 4 while a roll of 5-5 would be given a value of 5).
Mathematics
1 answer:
Sati [7]3 years ago
7 0

Answer:

4.4722

Step-by-step explanation:

The following table shows the 36 elements of the sample space of the experiment

\left[\begin{array}{cccccccc}&1&2&3&4&5&6\\--&--&--&--&--&--&--\\1\;|&1&2&3&4&5&6\\2\;|&2&2&3&4&5&6\\3\;|&3&3&3&4&5&6\\4\;|&4&4&4&4&5&6\\5\;|&5&5&5&5&5&6\\6\;|&6&6&6&6&6&6\end{array}\right]

From this table we can compute the probabilities :

P(X = 1) = 1/36

P(X = 2) = 3/36

P(X = 3) = 5/36

P(X = 4) = 7/36

P(X = 5) = 9/36

P(X = 6) = 11/36

So the expected value equals

1P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)+5P(X=5)+6P(X=6) =

= 1/36 + 6/36 + 15/36 + 28/36 + 45/36 + 66/36 = 161/36 =  

4.4722

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SVEN [57.7K]
In this type of calculations, we decompose 13 by checking the lowest powers of the base, that is 40. for example we check 40^2, or 40^3 and compare it to 85

Notice

40*40*40=64,000

so we check how many time does 85 fit into 64,000:

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85*753=64,005;       64000-64,005=-5

this means that 

40^{3} =-5\pmod{85}

thus

40^{13} =40^{3*4+1}={(40^{3})}^{4}*40=(-5)^{4}*40 \pmod{85}=\\\\625*40\pmod{85}=(7*85+30)*40\pmod{85}=30*40\pmod{85}\\\\=1200\pmod{85}=(14*85+10)\pmod{85}=10\pmod{85}


Answer: 10 (mod85)

Remark, the set of all solutions is:

{......-75, 10, 95, .....}, that is 85k +10
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