While playing a certain dice game, Chris wins if the sum of the two dice is 10, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?  Choose the correct answer:  A) 1/2  B) 17/36  C) 190/827  D) 11/36  E) 25/216
Probability of winning game = 1- Probability of losing game   Probability of losing game = (Probability of not getting sum 7 in any of the three attempts)   Ways of getting sum 10 = (4,6)(5,5)(6,4)= 3 ways   Total ways of getting outcome on two dice =6*6= 
<gadget id="calculator">6*6</gadget>
<output>36</output>
36   Probability of getting sum as 10 in any attempt=3/36=1/12   Probability of NOT getting sum as 10 in any attempt = 1-(1/12)= 11/12   Probability of losing game =(11/12)*(11/12)*(11/12)=637/827   I.e. Probability of wining game = 1-(637/827) = 190/827   Answer : Option 190/827
<result>C</result>
While playing a certain dice game, Chris wins if the sum of the two dice is 10, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?