Twentytwo slips of paper each marked with a different letter

Twenty-two slips of paper each marked with a different letter of the alphabet and placed in a basket. A slip is pulled out, its letter recorded (in the order in which the slip was drawn), and the slip is replaced. This is done 3 times. Find the probability that a word with no repetition of letters is formed.

The probability is __ ( type an integer or a simplified fraction.)

Solution

no of papers=22

since each paper consist of single unique alphabet then no of alphabet used=22

probability of first slip=1/22

then slip is replaced but word should have no repeated letter

When we sample with replacement, the two sample values are independent. Practically, this means that what we get on the first one doesn\'t affect what we get on the second. Mathematically, this means that the covariance between the two is zero.

then probability of second slip=1/22

probability of third slip=1/22

total probability = 1/22*1/22*1/22=0.000091125