. Assume an automobile license plate number consists of two letters followed by a three digit number. How many distinct license plate numbers can be formed if
(a) there are no restrictions and
(b) the letters O and I are not used?
(c) What is the probability of selecting at random a license plate that ends in an even number?
A)
26^2 times 10^3 = 676,000
B)
24^2 times 10^3 = 576,000
C)
All plates end in a digit 0-9, half of which are even.
You have a 1/2 probability of getting an even one at random.
December 4th, 2009 at 3:26 am
A)
26^2 times 10^3 = 676,000
B)
24^2 times 10^3 = 576,000
C)
All plates end in a digit 0-9, half of which are even.
You have a 1/2 probability of getting an even one at random.
References :
December 4th, 2009 at 3:53 am
Well, let’s think about how many different two digit numbers can be formed. We know there are 100 possibilities between 00 and 99, but how can we use that to figure out how many different letter schemes we can use. This is an example of counting theorem (or something like that).
For the first digit, there are 10 distinct numbers to use, and for the second digit, there are 10 distinct numbers to use. So for the two digit system, there are 10*10, or 100 distinct two digit numbers. So you see that the total number of possibilities depends on the number of choices you have for each individual slot.
For (a), the total number of distinct license plate numbers that can be formed are:
26*26*10*10*10
Where the 26*26 describes how many different ways we can make a two letter system and 10*10*10 describes how many ways we can describe a three digit system.
For (b), you do the same thing. However, those two letters cannot be used in either of the spaces, so instead of 26*26, you get 24*24.
For (c), if you want to get a license plate that ends in an even number, you need not consider anything in the first four spaces. After all, all we care about is the last slot. So what is the probability of the last slot being a 0, 2, 4, 6, or 8? There are 10 different numbers to choose from, and we’re hoping to get one of those 5 numbers, so the probability is:
5/10 = 50%.
References :