Question

A locker can be opened by dialing a fixed three-digit code (between 000 and 999). A stranger, who does not know the code, tries to open the locker by dialing. The stranger succeeds at the\[{k^{th}}\], finding 1000 p. (Assume that the stranger does not repeat unsuccessful combinations.)

Answer 1

Hint: The event is the outcome of any trails. Let us take the die as an example, a die has 6 sides and when it is rolled, the outcomes will be 1, 2, 3, 4, 5, and 6. Then roll the dice is called the trial and the outcome of the rolled dice is called as the event. So assume p be the event in the solution. The combinations are defined as the arrangement of the things or items in different ways.

Complete step-by-step answer:
The number of digits in the locker dialing is 3-digit.
The limits of the three digits are from 000 to 999.
Let the event that the stranger can succeed at a particular trial is P(A).
The equation to find that the stranger success in his trials is,
\[P(A) = \dfrac{{999}}{{1000}} \times \dfrac{{998}}{{999}} \times \dfrac{{997}}{{998}} \times .... \times \dfrac{{1000 - k + 1}}{{1000}} \times \dfrac{1}{{1000 - k + 1}}\]
Then by solving the above expression, we will get
\[\begin{array}{c}
P(A) = \dfrac{1}{{1000}}\\
 = 0.001
\end{array}\]
Now we have to find the value of 1000p then, substituting the value of p in the term, we will get
\[\begin{array}{c}
1000p = 1000\left( {0.001} \right)\\
 = 1
\end{array}\]
Therefore, the value of the 1000p is 1.

Note: In the solution, be careful while solving the equation and it should be noted that the stranger has to trial the codes that are not repeated once. For example, if a stranger trialed code 555, then he should trail the same combination again.