<s>[INST] The probability of getting a 7 This problem is probably too easy for all you math geniuses out there, but I am having trouble understanding it, so here it is:

An athlete has 10 cards, numbered 1 to 10. Every day, he picks a card randomly and does that number of pushups. He doesn't replace the card, and picks until all the cards are gone. If after a certain number of tries, the athlete has done a total of 12 pushups, then the probability of him/her picking a 7 as the very next card is $\frac{a}{b}$. What is a + b?

I know that I have to count the possibilities of getting a 12:


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*2 cards
The possibilities would be 2 and 10, 3 and 9, 4 and 8, 5 and 7, which would make 4 possibilities.


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*3 cards
1, 2, 9 | 1, 3, 8 | 1, 4, 7 | 1, 5, 6 | 2, 3, 7 | 2, 4, 6 | 3, 4, 5 | which would be 7 possibilities.


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*4 cards
1, 2, 3, 6 | 1, 2, 4, 5 | which is 2 possibilities.
There cannot be more than 4 cards: 1, 2, 3, 4, 5 would make 15.
The total number of possibilities for getting a 12 is 17, and out of those, there are 3 of them that will pick the card 7, which means 14 out of the 17 possibilities will be valid for the problem.
I don't know how to continue from here. Should I solve for the probability each of the three cases listed above separately, then add them, or is there a better method?
 [/INST] First calculate the relative probabilities of getting a $2, 3$ or $4$ card sum of $12$.
These are:$$\frac{4}{\binom{10}{2}};\frac{7}{\binom{10}{3}}; \frac{2}{\binom{10}{4}} = \frac{4}{45}; \frac{7}{120}; \frac{2}{210} = \frac{448}{5040}; \frac{294}{5040}; \frac{48}{5040}$$
Given the sum is $12$, the actual probabilities of being a $2, 3$ card or $4$ card sum are:
$$\frac{448}{790}; \frac{294}{790}; \frac{48}{790}$$
The sum of all $3$ probabilities of drawing a $7$ from each of the $3$ card quantities is therefore:
$$P(7) = \frac{448}{790}\cdot \frac{3}{4}\cdot \frac{1}{8} + \frac{294}{790}\cdot \frac{5}{7}\cdot \frac{1}{7} + \frac{48}{790}\cdot \frac{1}{6} = 0.101266$$
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