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 anshu
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...
6 years ago
Abhishek   c)
6 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

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