Rohan and Vikas decide to play a game which involves flipping of a coin. In a round, Rohan gives rs 5 toVikas if the coin shows tail whereas Vikas gives rs 2 to Rohan if the coin shows head. They start the gamewith a biased coin in which head is likely to occur 50% more times than tail If initially, each of the two has 50, then what is the probability that at the end of 10th round Vikas would have at least 80 rupees in all?
Rohan and Vikas decide to play a game which involves flipping of a coin. In a round, Rohan gives rs 5 toVikas if the coin shows tail whereas Vikas gives rs 2 to Rohan if the coin shows head. They start the gamewith a biased coin in which head is likely to occur 50% more times than tail If initially, each of the two has 50, then what is the probability that at the end of 10th round Vikas would have at least 80 rupees in all?
probability of coin showing tails: 2/5, heads:3/5 Since vikas needs at least 80 rupees, it can happen only when he wins in>=8 rounds. So case1: he wins in 8 rounds and loses other 2: (10c8*2^8*3^2)/5^10
case 2: he wins in 9 rounds and loses 1 round: (10c9*2^9*3^1)/5^10
case 3: he wins all 10 rounds: (10c10*2^10*3^0)/5^10
add them all up to get required probability as (469*2^8)/5^10
Rohan and Vikas decide to play a game which involves flipping of a coin. In a round, Rohan gives rs 5 toVikas if the coin shows tail whereas Vikas gives rs 2 to Rohan if the coin shows head. They start the gamewith a biased coin in which head is likely to occur 50% more times than tail If initially, each of the two has 50, then what is the probability that at the end of 10th round Vikas would have at least 80 rupees in all?
In how many ways can two kings be placed on a regular chessboard, such that none of them attacks the other?
Not possible since 2 kings cannot attack each other at all..it is considered illegal acc to chess rules to move a king to a square next to another king
Positions when they can attackCase 1 ) One King placed at corner = 3 positions of attack => 4*3 = 12 ( 4 corners) Case 2) One king on the sides ( except corners ) = 5 ways => 24*5 = 120 ( 6 squares each side = 24 positions )Case 3) Some where in the middle = 8 positions of attack => 36*8 = 288 ( 6*6 squares in the middle) So total cases when attack can happen on any one = 12+120+288 = 420 So non attack position = 64C2 (any two positions for the 2 kings) - 420 = 1596
And considering they are black and white, they can switch places too so 1596*2 = 3192....
Not possible since 2 kings cannot attack each other at all..it is considered illegal acc to chess rules to move a king to a square next to another king
I think Chess is very close to your heart so you can't see any of its rules getting violated(the case is same with me also) But you can keep it simple and take it as a question and solve it by considering no such restrictions (taking the case where a king can move in any direction only one step at a time)
I think Chess is very close to your heart so you can't see any of its rules getting violated(the case is same with me also)But you can keep it simple and take it as a question and solve it by considering no such restrictions (taking the case where a king can move in any direction only one step at a time)
Yeah..when i read the question i was flummoxed since i had always been taught that 2 kings can never come within 1 square of each other
