# University Combinatronics (Enumeration) and Basic Probability

• Sep 15th 2010, 01:09 AM
aarnold
University Combinatronics (Enumeration) and Basic Probability
Hi! First post here.

This is actually for a university level course in Discrete Mathematics. This assignment question regards the topic of enumeration. I'm pretty terrible at basic probability, so I was hoping someone could check my answers for me. I'm not feeling so confident.

Okay, the problem:

Consider a combination lock for a bicycle. It has four tumblers, each with the digits 0-9. The rider is lazy and only twists two (random) tumblers to lock the bike. What is the probability of guessing the correct combination.

Without knowledge of which tumblers have been twisted, we must randomly choose two from the four. This is an unordered selection without repetition.

The formula for the number of unordered selections without repetition of r objects from an n-set is n choose r.

If n = 4 and r = 2, then the number of selections = (4 choose 2) = 6

So there is a 1/6 = 0.1667 chance of selecting the correct tumblers to twist.

The formula for the number of ordered selections with repetition of r objects from an n-set is nr.The probability of selecting the correct two digits on the tumblers is therefore 1/102= 1/100 = 0.01

The selection of the digits is not dependent on the selection of the tumblers. However, to correctly guess the combination the correct digits must be applied to the correct tumblers. So therefore the probability of randomly guessing the correct combination is:

Probability of guessing which two tumblers to turn
multiplied by
Probability of guessing which digits to turn to

0.1667 * 0.01 = 0.001667

The main part I'm unsure about is the last bit. Any feedback would be great. Thanks!

• Sep 15th 2010, 01:30 AM
Traveller
Seems okay.