Ahmed is playing a lottery game where he must pick 2 numbers from 0 to
9 and then one letter out of the 26-letter English alphabet. He may
choose the same number both times. If his ticket matches the 2 numbers
and 1 letter drawn in order, he wins the grand prize and receives
\$10,405.If just his letter matches but one or both of his numbers do not
match, he wins the small prize of \$100. Under any other outcome he
loses and receives nothing. The game costs him \$5 to play. He has
chosen the ticket 04R. Assuming that he paid the \$5 and he picked the
ticket 04R.
The solution to the problem is found in this video: https://www.youtube.com/watch?v=6vlBOHckmzU&feature=youtu.be
In the video, the probability of getting the small prize is (1/26) – (1/2600). Shouldn't the probability of winning the small prize be the sum of the probabilities of getting the letter right and 1 number wrong + getting the letter right and 2 numbers wrong: (1/26)(9/10)(1/10) + (1/26)(9/10)(9/10)?
Best Answer
Probability of getting one letter right and 1 number wrong isn't $\frac{1}{26}\cdot\frac{9}{10}\cdot\frac{1}{10}$, it's twice this. After this fix both answers are the same.
It can be easier to see in smaller and simpler problem. For example, if you toss two biased coins with probability of head $\frac{1}{3}$ and of tail $\frac{2}{3}$, probability of getting one head and one tail isn't $\frac{1}{3} \cdot \frac{2}{3}$ - it's probability of, say, "the first coin lands head, the second lands tail" (A), or vice versa (B). And the event "one coin lands head, one coin lands tail" is (disjoint) union of (A) and (B) - so it's probability is sum of probabilities of (A) and (B).