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The probability of any event is defined as the ratio of the number of outcomes associated with the event to the total number of possible outcomes. The probability of a particular event is always less than or equal to 1.

A three-digit number lies from and thus, there are 900 three-digit numbers. So, the sample space for the given problem is all the three-digit number from 100 to 999, that is expressed as follows,

$100鈥�鈥�,鈥�鈥�101鈥�鈥�,鈥�鈥�鈥�鈥�鈥�鈥�鈥�,鈥�鈥�999$

Each digit of the obtained sample space has an equal probability of $\frac{1}{900}$.

A three-digit number which end with 7 are 90 namely$\left(107,117,鈰�,197,207,鈰�,297,鈰�,907,917,鈰�,997\right)$

with each having a probability of$\frac{1}{900}$ .

Find the probability that the units digit is 7 by adding the probabilities of each possible outcomes, that is 90 times $\frac{1}{900}$.

$\begin{array}{rcl}p& =& 90脳\frac{1}{900}\\ & =& \frac{1}{10}\\ & & \end{array}$

Thus, the probability that the units digit is 7is$\frac{1}{10}$ .

A three-digit number which starts with 7 are 100that are$\left(700,701,鈰�,799\right)$ with each having a probability of $\frac{1}{900}$.

Find the probability that the hundreds digit is 7by adding the probabilities of each possible outcomes, that is 100 times$\frac{1}{900}$ .

role="math" localid="1655788370190" $\begin{array}{rcl}p& =& 100脳\frac{1}{900}\\ & =& \frac{1}{9}\\ & & \end{array}$

Thus, the probability that the hundreds digit is 7 is $\frac{1}{9}$.