When four digits (between 0 and 9 inclusive) are randomly selected with replacement for a lottery ticket, what is the probability that at least one digit will be a zero?

We are asked to determine the probability of having at least one digit of zero in a winning lottery number that is composed of four digits from the set of numbers from 0 to 9. To determine the probability to be known, we can approach this problem by computing for the probability by counting theory for each digit.

Digit 1: there are 10 choices, so the probability is 1/10
Digit 2: there are only9 choices, probability of 1/9
Digit 3: there are also 9 choices, probability of 1/9
Digit 4:  9 digits as 0 is not a included

*(numbers beginning with zero do not count zero as part of the number)

Hence the probability is (1/10)*(1/9)*(1/9)*(1/9) = 0.000137

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post