0

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=

Joffan
  • 39,627
laki
  • 1
  • Hi, I broke up your text a bit for readability and highlighted the question. Are you trying to find the inverse case? or did you just lose track of which letter was specified to start? – Joffan Jun 18 '15 at 16:26

1 Answers1

0

$$n(S) = 2 \times 10 \times 10 \times 10 \times 26$$

$$n(A) = 1 \times 10 \times 10 \times 5 \times 21$$ ( assuming "y" is not a vowel)

so $$P(A) = \frac{n(A)}{n(S)} = \frac 12 \times \frac{5}{10} \times \frac{21}{26} $$

WW1
  • 10,497