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 A and B play a game between them. The dice consist of colors on their faces(instead of number). When the dice are thrown, A wins if both show the same color, otherwise B wins. One die has 3 red faces and 3 blue faces. How many red and blue faces should the other die have if the both players have if the both players have the same chances of winning? a) 5 red and 1 blue faces. b) 1 red and 5 blue faces. c) 3 red and 3 blue faces. (TCS Question) Asked by anshuCompile Program
 Ranendra kumar   c) 7 years ago
 Gautham Swami   According to the question, For both of them to have an equal chance of winning, The probability that a single color pops up on the two dice should be half So, Probability (dice 1 is red AND dice 2 is red) + Probability (dice 1 is blue AND dice 2 is blue) = 1/2 Let the other dice have x red faces So, it will have 6-x blue faces So, (4/6)*(x/6) + (2/6)*(6-x)/6 = 1/2 So, x = 3 Hence, Option C is the answer. 7 years ago
 Nitin Meena    answer should be c b... 7 years ago
 Abhishek   c) 7 years ago
 rakesh verma   it is independent on the no. of red and blue side of the other coin winning probability=(1\2)*(x/6)+(1/2)*(6-x)=1/2 ny value of x satisfies this equ 6 years ago
 vikas gupta   Ans..C because prob. of given 1st die is 1/2,so for 2nd die prob. must should be 1/2.... 6 years ago
 Subhajit Pramanik   c 6 years ago
 ratikant patra   c 6 years ago
 sushanth   @gautham swami can u plz xplain why did we take 4/6 and 2/6. 4 years ago
 sasidhar kumar   option 3 is correct equal chance to winning 4 years ago