I was studying for some quizzes when a wild question appears. It goes like this:
If each coded item in a catalog begins with 3 distinct letters followed by 4 distinct nonzero digits, find the
probability of randomly selecting one of these coded items with the first letter a vowel and the last digit
even.
My work:
If I were to pick a random catalog, I might pick a catalog having the name $ABC \space 1234$. Moreover, if I were to arrange all the known combinations of the catalog having 3 distinct letters followed by 4 distinct nonzero digits, the $ABC \space 1234$ is the first to be listed. I will never see such thing as $ABC \space 1233$, because that would violate the rule which states the the catalog must begin with 3 distinct letters followed by 4 distinct nonzero digits.
With that in mind, the probability ($P$) of finding a catalog having these attributes (first letter a vowel and the last digit
even) would be:
$$P = \frac{number\space of\space catalogs\space having \space a \space first \space letter\space a \space vowel \space and \space the \space last \space digit \space even}{total \space number\space of\space catalogs\space possible \space with \space 3 \space distinct \space letters \space followed \space by \space 4\space distinct \space nonzero\space digits}$$
For the first part:
Since there are 5 vowels in an English alphabet, there are 5 choices for the first letter. For the second letter, assuming that one of the vowels
was used on the first letter, there are 25 remaining choices for the second letter. For the third letter, assuming that one of the letters
was used on the second letter, there are 24 remaining choices for the second letter.
For the second part:
Since the last digit must be even, there are 5 choices for the last digits (0,2,4,6,8). For the first number, assuming that one of even numbers was used
on the 4th digit (let's say it's $2$), there are 9 choices (0,1,3,4,5,6,7,8,9). For the second number, assuming that one of the numbers was used
on the 1st digit (let's say it's $1$), there are 8 choices (2,3,4,5,6,7,8,9). For the third number, assuming that one of the numbers was used
on the 2nd digit (let's say it's $3$), there are 7 choices (2,4,5,6,7,8,9).
The total number of catalogs with 3 distinct letters followed by 4 distinct nonzero digits would be $(26)(25)(24) \space (10)(9)(8)(7)$ catalogs.
Making the mathematical expression to get what I mean:
$$P = \frac{(5)(25)(24) \space (9)(8)(7)(5)}{(26)(25)(24) \space (10)(9)(8)(7)}$$
Ultimately, the probability I get is $\frac{5}{52}$.
My books's answer was $\frac{10}{117}$.
What erroneous idea did I used that made me lead to my wrong answer? How do you answer the above problem?
Best Answer
You can ignore all the digits and letters except the first and last. This shows up in the fact that the $25,24,9,8,7$ all cancel. You have five vowels out of $26$ letters and four nonzero even digits out of nine. $\frac 5{26} \cdot \frac 49=\frac {10}{117}$. You missed the nonzero part. The chance from $\frac 5{10}$ to $\frac 49$ for the even digit accounts for the difference between your answer and the book's.