Calculate These Two squares are chosen at random on 8 x 8 chessboard. What is th

Question

Calculate These Two squares are chosen at random on 8 x 8 chessboard. What is the probability that they share a side? 8 rooks are placed randomly on an 8 x 8 chessboard. What is the probability none of them are attacking each other? (Two rooks attack each other if they are in the same row, or in the same column). A bag has two quarters and a penny. If someone removes a coin, the Coin-Replenished will come and drop in 1 of the coin that was just removed with 3/4 probability and with 1/4 probability drop in 1 of the opposite coin. Someone removes one of the coins at random. the Coin-Replenished drops in a penny. You randomly take a coin from the bag. What is the probability you take a quarter?

Explanation / Answer

(c) Two squares are chosen at random .Probability that they will share a side.

The number of permutation to choose 2 squares = 64C2 = 2016

Now, we will calculate the number of squares who are adjacant to each other. In * X * chess board, there are 3 different type of squares.

4 are the corner squares who have only 2 sides so number of squares's who have common sides = 4 * 2

6 * 4 = 24 are the edge squares who have only 3 sides so number of squares's who have common sides. = 6 * 4 * 3

36 are the squares in middle who have 4 sides so number of squares's who have common sides. = 36 * 4

so total number of squares pair = (4 * 2 + 6 * 4 * 3 + 36 * 4 )/2 = 112

so total squares are who are adjacent to each other = (4 * 2 + 6 * 4 * 3 + 36 * 4 )/2016 =112/2016 = 0.0555

(d) 8 rooks are randomly placed on 64 squares, so number of permutation for rooks to be there = 64C8

so here it is asked that no rooks will attack the other rook so no rook should be in the same row or in the same column of the other rook.

so for the first rook there are 82 possibility to be there in the chess board , but if the first rook is placed anywhere in the chess board it will block ( 8+8-1 = 15) sites for the second rook, s second rook will have only 72 position left and so on.

so number of possibilities tfor rooks that they will not attack each other = 82 * 72 * 62 * 52 * 42 * 32 * 22 * 12

so Probaiblity that no rook will attacjk the other rook = (82 * 72 * 62 * 52 * 42 * 32 * 22 * 12 )/64C8 = 1625702400/4426165368 = 0.3672

(e) Bag has 2 quarteres and a penny. Here we have to calculate thatafter one attempt of removing a coin, and its replenishment, what will be the probability that i will draw a quarter.

P ( quarter in later draw) = P ( quarter in first draw) * P( quarter in second draw) + P( penny in first draw) * ( quarter in second draw)

we will do this question by evaluating both terms one by one

(1) If we pick quarter in first draw which have the probability of 2/3, so it is 3/4 chance that new replnished coin is a quarter and 1/4 chance that the replinshed one is a penny.so getting a quarter in second draw, if quarter is replenished will be 2/3 and if penny is replnished will be 1/3 so here in first case the probability is =

2/3 * 3/4 * 2/3 + 2/3 * 1/4 * 1/3 = 7/18

(2) If we pick penny in first draw which have the probability of 1/3, so it is 3/4 chance that new replnished coin is a penny and 1/4 chance that the replinshed one is a quarter .so getting a quarter in second draw, if quarter is replenished will be 2/3 and if penny is replnished will be 1/3 so here in first case the probability is =

2/3 * 1/4 * 2/3 + 2/3 * 3/4 * 1/3 = 5/18

so probability of getting getting quarter in later draw = 7/18 + 5/18 = 2/3, which is equal to the probability of drawing a quarter in first draw.

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