Twelve people at a sale all want to buy a TV, but there are only two left. The TVs are allocated at random. What is the probability that one particular person will get one of the TVs?

Is it 2/12 or is it 1/12 + 1/11 ? Or something different ?

Let's be pessimistic. What's the probability you don't get the first one? Then, presumably there are only 11 people still in play and ten of them will be unlucky. What's the probably you don't get the second one?

Now we need to multiply these probabilities, because we think we _won't_ get the first one AND we _won't_ get the second one.

This gives you the probability you will be unlucky and get neither TV. Now can work out the probability you do get one.

Reconsideration of my earlier : does depend on selection method. If say 12 tickets with 2 drawn from hat, then to win needs: (win Ticket1) OR (notwin ticket 1 and win ticket 2) , so

1/12 +11/12 x 1/11 = 2/12

However I agree 'pessimistic' method is more elegant!

I did O level maths (before GCSE) followed by A level Pure and Applied, and then into a science degree, without learning any statistics! Where would this probability topic fall in the current curriculum? Is it GCSE or A level?