Question

You are playing a solitaire game in which you are dealt three cards without replacement from a simplified deck of 10 cards (marked 1 through 10). You win if one of your cards is a 10 or if all of your cards are odd. How many winning hands are there if different orders are different hands? and also what is your chance to win

Answer 1

We have that:

If the first card drawn is a 10, then there are $P_9^2 = \frac{9!}{(9 - 2)!} = 9\cdot 8 = 72$ hands;

If the second card is a 10 then there are 9 possible numbers for the first card and 8 possible numbers for the third card, it gives us $9 \cdot 8 = 72$ hands;

If the third card is 10 then there are 9 possible numbers for the first card and 8 possible numbers for second card, it gives us $9 \cdot 8 = 72$ hands;

And there are 5 odd cards, and it gives us: $P_5^3 = \frac{5!}{(5 - 3)!} = 5\cdot 4\cdot 3 = 60$ hands.

So, we have that in total we have $72 + 72 + 72 + 60 = 276$ winning hands.

In total we have: $P_{10}^{3} = \frac{10!}{(10 - 3)!} = 10\cdot 9\cdot 8 = 720$ possible hands. So, the chance that we will win is

$$\frac{276}{720} \approx 0.383.$$

Answer: total number of winner hands is 276. The chance that I will win is $\frac{276}{720} \approx 0.383$.