# B OR probabilities

1. Apr 28, 2016

### little neutrino

A code is generated consisting of 3 letters followed by 2 digits. Each of the 3 letters generated is equally likely to be any of the 26 letters of the alphabet. Each of the 2 digits generated is equally likely to be any of the nine digits 1-9. Find the probability that a randomly chosen code has exactly 2 letters the same or 2 digits the same, but not both.

Attempted solution:

P(exactly 2 digits same) = (263*9)/(263*92) = 1/9
P(exactly 2 letters same) = (26*25*3*92)/(263*92) = 75/676
P(2 digits same AND 2 letters same) = (26*25*3*9)/(263*92) = 25/2028
Required probability = 1/9 + 75/676 - 25/2028 = 0.210

But the correct answer is 0.197. Can I ask where did I go wrong? I can't figure it out...
Thanks!

2. Apr 28, 2016

### stevendaryl

Staff Emeritus
Your calculation for 2 letters is the probability that at LEAST 2 letters is the same. You have to subtract off the probability that all 3 letters are the same to get the probability that exactly 2 letters are the same.

3. Apr 28, 2016

### cpscdave

One other possible mistake. Exactly how many ways are there to have 2 items the same in a set of 3?

4. Apr 28, 2016

### little neutrino

But why is that so?
__ __ __ __ __
For two letters to be the same == 26 choices
For the last letter, since it can't be the same as the two letters that are the same == 26 - 1 = 25 choices
No. of ways to arrange 3 letters, when 2 are the same == 3!/2! = 3
Thus no. of ways == 26 * 25 * 3 * 92

Why did I calculate the probability that at least 2 letters are the same? Could you elaborate? Sorry kind of confused.

5. Apr 28, 2016

### cpscdave

Yup I'm dumb sorry :D

6. Apr 28, 2016

### stevendaryl

Staff Emeritus
Hmm. My criticism was mistaken, but here's a suggestion: Instead of computing probabilities for the various possibilities, try just computing

$N_L =$ the number of ways to have exactly 2 letters the same, and both digits different.
$N_D =$ the number of ways to have 3 or 0 letters the same, and both digits the same.

Then the probability is $\frac{N_L + N_D}{26^3 \cdot 9^3}$

When I did it that way, I got the expected answer.

7. Apr 28, 2016

### stevendaryl

Staff Emeritus
Okay, I figured out what you did wrong:

You used $P(A \vee B) = P(A) + P(B) - P(A \wedge B)$ where $A$ = "exactly 2 letters the same" and $B$ = "both digits the same". But they aren't asking for $A \vee B$, they're asking for $A\ xor\ B$ ($xor$ = exclusive or). The formula for that is:

$P(A\ xor\ B) = P(A) + P(B) - 2P(A \wedge B)$