the number plate on a car

Question

In a certain country, the number plate on a car consists of any 3 letters of the alphabet (the first letter is always a "K" or a G), followed by any 3 digits (0 to 9) and a alphabet. For a car chosen at random, what is the probability that the number plate starts with a "G" and ends with an odd digit and a non-vowel? The number plate starts with a ’K’, so there is only 1 choice for the first letter, and ends with an even digit, so there are 5 choices for the last digit (1,3,5,7,9), 5 vowels and 10 possible choices for each of the other digits. Therefore

Number of events=?

Total number of possible number plates=?

Probability=?

Answer 1

I'm guessing that there are seven digits in the license plate. 

 

if A stands for any letter, and # is a number, we'd have KAA###A or GAA###A.  We can look at each digit as an independent event, since each has no effect on the probability of the others being what they are.  Furthermore, we need to know the probability of the first digit being a G or a K.  Although it's not likely (can you think of why?), we can assume these to be equal (1/2). 

 

For independent events, the total probability is the product of all probabilities taken together.  Let's call the first digit/letter D1, the second to last D6, and the last D7.  What we have is

P(D1 = K and D6 = odd and D7 = non-vowel) = P(D1 = K) * P(D6 = odd) * P(D7 = non-vowel).

 

Let's work out each probability:

P(D1 = K) = 1/2

P(D6 = odd) = 5/10 = 1/2

P(D7 = non-vowel) = 21/26.

 

So the probability of all of these events occurring on the same license plate is (1/2)*(1/2)*(21/26), or .202 (20.2%).