Question

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9). Uppercase letters are not used. Let A denote the event that a password begins with a vowel (either a, e, i, o, or u), and let B denote the event that a password ends with an even number (either 0, 2, 4, 6, or 8). Suppose a hacker selects a password at random. Determine the following probabilities: a. P(A) b. P(B) c. P(A ∩ B) d. P(A ∪ B)

Answer 1

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$

$exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8$

$ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8$

$oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8$

$uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8$

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

$P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}$

Finally, you can compute the probability of the union using the formula

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Since we already computed all these quantities.