ANotSmartPerson

Thanks!

Thank you for trying but apparently the answer was wrong, here is the solution I got back:

We consider the opposite; we try to find the number of words that do not contain A, and then subtract it from the total possible number of words. So we have a few cases to consider:

 One letter words: There is only 1 one-letter word that contains A, that's A.

 Two letter words: There are 19*19=361 words that do not contain A. There is a total of 20*20=400 words, so we have 400-361=39 words that satisfy the condition.

 Three letter words: There are 19*19*19=6859 words without A, and there are 20^3=800 words available. So there are 1141 words that satisfy the condition.

 Four letter words: Using the same idea as above, we have 20^4-19^4 words satisfying the requirement.

So this gives a total of 1+39+1141+29679=30860 words.

Thanks :)

Thanks!

Thank you for trying, but the answer was incorrect :(  Here is the solution I got back in case it helps. 

We approach this problem by trying to find solutions to the equation pq-4p-2q=2 . To do this, we can use Simon's Favorite Factoring Trick and add 8 to both sides to get qp-4p-2q+8=10. This can be factored into

  (p-2)(q-4)=10

We now can see that there are solutions only if p-2 divides 10 . Thus,  there are 4 possible values of p between 1 and 10 inclusive (1,3,4,7) . It follows that the probability of picking such a p is 2/5.

Thanks again!

79+124+90=293 360-293=67? I'm not sure, you might not want to trust this answer.

Thanks for helping

Thank you, that was fast!