What are your chances of winning in the following scenario?
Consider a lottery game with 49 numbers and 6 are chosen as the winning numbers. You're allowed to select 10.
What is the probability that your selection contains all 6 winning numbers? (the remaining 4 can be anything else)
Please confirm if this is correct:
${43 \choose 4}$ / ${49 \choose 10}$
Reasoning:
From a selection of 10 numbers, there are ${43 \choose 4}$ combinations which contain the winning 6; i.e. ${6 \choose 6}$ * ${43 \choose 4}$
divided by the total number of combinations with 10 numbers: ${49 \choose 10}$
Going by what your question says, the remaining $4$ tickets don't matter, what is important is the winning $6$ tickets. And since you have all $6$ tickets, you are definitely going to win. So probability is $1$.
Hope this helps you.