A printing machine, capable of only printing the 10 digits, prints the right digit 12% of the time. Whenever the machine prints erroneously, all wrong possibilities are equally likely.

If a random key is pressed twice and a same digit comes out both times, what is the probability that it was the correct one?

We get a wrong first number 88% of the time.
We get a wrong second number 88% of the time, but that second number only matches  the first number 88/9 % of the time.
However, we can get 9 wrong numbers.

Thus, we are wrong .88 * .88 / 9 * 9 = .88^2 = .7744.
And following that, we are correct 1-.7744, or .2256.

22.56% of the time we get the right answer if both numbers are the same, substantially more than e.g. says.

Posted by Caz on 2006-07-22 20:30:47