Question

Two squats are chosen at random on a chessboard. The probability that they have a side in common is

Solution

Correct option is

1/18

The number of ways of choosing the first square is 64, and that for the second square is 63. Therefore, the number of ways of choosing the first and second squares is 64 × 63 = 4032. Now we proceed to find the number of favorable ways. If the first square happens it be any of the four squares in the corner, the second square can be chosen in 2 ways. If the first square happens to be any of the 24 (non - corner) squares on either side of the chessboard, the second square can be chosen in 3 ways. If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways. Therefore, the number of favorable ways is (4) (2) + (24) (3) + (36) (4) = 224. Thus, the probability of the required event is 224/4032 = 1/ 18.

SIMILAR QUESTIONS

Q1

A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is 

Q2

A bag contains a white and b black balls. Two players, A and B alternate ydraw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game. A begins the game. If the probability of A winning the game is three times that of B, the ratio ab is

Q3

A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the determinant chosen is non – zero is

Q4

A fair coin is tossed 100 times. The probability of getting tails an odd number of times is

Q5

One hundred identical coins, each with probability of head are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, the value of p is

Q6

Suppose X follows a binomial distribution with parameters n and p, where 0 < p < 1. If P (X = r)/P(X = n – r) is independent of n for every value ofr, then 

Q7

The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8 is

Q8

For the three events AB and CP(exactly one of the events A or Boccurs) = P(exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events AB and C occurring is

Q9

Nine identical balls are numbers 1, 2,…9. Are put in a bag. A draws a ball and gets the number a. the ball is put back the beg. Next B draws a ball gets the number b. The probability that a and b satisfies the inequality a – 2b + 10 > 0 is