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+0  
 
 #1
avatar+615 
-3

sí C-Phill está respondiendo (:

smiley

 Apr 10, 2019
 #2
avatar+101103 
+2

This is a binomial probability

 

P( no 3's)   =  (5/6)^10

 

P( one 3)  =  C(10, 1) * ( 1/6) (5/6)^9

 

P( two 3's)  = C(10, 2) *(1/6)^2 (5/6)^8

 

P( three 3's)  = C(10, 3) *(1/6)^3 (5/6)^7

 

P(four 3's) = C(10, 4) * (1/6)^4 (5/6)^6

 

So....the sum of these  ≈ .9845  ≈   98.45 %

 

 

cool cool cool

 Apr 10, 2019
 #3
avatar+615 
-4

Niza un C-Phill

Nickolas  Apr 10, 2019

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