Question

A certain three​-cylinder combination lock has 60 numbers on it. To open​ it, you turn to a number on the first​ cylinder, then to a second number on the second​ cylinder, and then to a third number on the third cylinder and so on until a three​-number lock combination has been effected. Repetitions are​ allowed, and any of the 60 numbers can be used at each step to form the combination.​ (a) How many different lock combinations are​ there? (b) What is the probability of guessing a lock combination on the first​ try? ​(a) The number of different three​-number lock combinations is nothing.

Answer 1

Answer:

(a) The number of different lock combinations is 216,000.

(b) The probability of getting the correct combination in the first try is $\frac{1}{216000}$.

Step-by-step explanation:

The lock has three-cylinder combinations with 60 numbers on each cylinder.

The procedure of the opening lock is to turn to a number on the first​ cylinder, then to a second number on the second​ cylinder, and then to a third number on the third cylinder.

The numbers can be repeated.

(a)

There are three cylinder combinations on the lock, each with 60 numbers.

It is provided that repetitions are allowed.

Then each cylinder can take any of the 60 numbers.

So there are 60 options for the first cylinder.

There are 60 options for the second cylinder.

And there are 60 options for the third cylinder.

So the total number of possible combinations is:

Total number of combinations = 60 × 60 × 60 = 216000.

Thus, the number of different lock combinations is 216,000.

(b)

The events of getting a correct combinations of the three​-cylinder combination lock implies that all the three cylinder are set at the correct numbers.

Each cylinder has 60 numbers.

This implies that there are 60 possible ways to get a correct number for the first cylinder.

The probability of getting the correct number for the first cylinder is:

P (Number on 1st cylinder is correct) = $\frac{1}{60}$.

Similarly for the second cylinder the probability of getting the correct number is:

P (Number on 2nd cylinder is correct) = $\frac{1}{60}$.

And similarly for the third cylinder the probability of getting the correct number is:

P (Number on 3rd cylinder is correct) = $\frac{1}{60}$.

So the probability of getting the correct combination in the first try is:

P (Correct combination in the 1st try) = $\frac{1}{60}\times \frac{1}{60}\times \frac{1}{60}=\frac{1}{216000}$.

Thus, the probability of getting the correct combination in the first try is $\frac{1}{216000}$.