GYanggg

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 #1
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Hey Guest! I'm not too familiar with expected value, but I can still give it a go:

 

NOTE: THIS SOLUTION IS WAY TO DIFICULT, LOOK AT MY SOLUTION TO 2. FOR THE CORRECT ONE. 

 

I'm not sure if this is the best way to do this, but it works. 

 

The possible sums of two dice are \((2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)\)

 

The probabilty of getting a sum of 2 is: \(\frac{1}{36}\). The numerator is one, because there is one way of getting a sum of 2, 1 + 1. 

 

We do the same for the rest of the sums, 

 

The probabilty of getting a sum of 3 is: \(\frac{2}{36}\), (1,2) ; (2,1)

The probabilty of getting a sum of 4 is: \(\frac{3}{36}\), (1,3) ; (3,1) ; (2,2)

The probabilty of getting a sum of 5 is: \(\frac{4}{36}\), (1,4) ; (4,1) ; (2,3) ; (3,2)

The probabilty of getting a sum of 6 is: \(\frac{5}{36}\), (1,5) ; (5,1) ; (2,4) ; (4,2) ; (3,3) 

The probabilty of getting a sum of 7 is: \(\frac{6}{36}\), (1,6) ; (6,1) ; (2,5) ; (5,2) ; (3,4) ; (4,3)

The probabilty of getting a sum of 8 is: \(\frac{5}{36}\), (2,6) ; (6,2) ; (3,5) ; (5,3) ; (4,4)

The probabilty of getting a sum of 9 is: \(\frac{4}{36}\), (3,6) ; (6,3) ; (4,5) ; (5,4) 

The probabilty of getting a sum of 10 is: \(\frac{3}{36}\), (4,6) ; (6,4) ; (5,5) 

The probabilty of getting a sum of 11 is: \(\frac{2}{36}\), (5,6) ; (6,5) 

The probabilty of getting a sum of 12 is: \(\frac{1}{36}\), (6,6)  

 

The expected value is: 

 

\((\frac{1}{36}\cdot2)+(\frac{2}{36}\cdot3)+(\frac{3}{36}\cdot4)+(\frac{4}{36}\cdot5)+(\frac{5}{36}\cdot6)+(\frac{6}{36}\cdot7)+(\frac{5}{36}\cdot8)+(\frac{4}{36}\cdot9)+(\frac{3}{36}\cdot10)+(\frac{2}{36}\cdot11)+(\frac{1}{36}\cdot12)\)

 

When you simplify: 

 

\(\frac1{18}+\frac16+\frac13+\frac59+\frac56+\frac76+\frac{10}{9}+\frac11+\frac{10}{12}+\frac{11}{18}+\frac13\)

 

When you add these up, you get 7. 

 

I hope this helped,

 

Gavin

GYanggg 20-may-2018 8:45:41
 
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