(ii) We have to arrange 4 cards in order, 2 of which are the same, and the remaining 2 different. There are ways of choosing the positions which the identical cards will occupy, and then 10 ways of choosing which card this will be. There are then 9 x 8 ways of arranging the two remaining cards around these two. Total: 6 x 10 x 9 x 8. So the probability of obtaining exactly three different cards is ...?
(iii) How many of the combinations will contain X but not Y? Y but not X? both X and Y? Add these numbers together, and then work out the probability that one of these combinations is chosen.
For the final part, once 9 different cards have been obtained, the probability that the next card completes the set is 0.1. So the probability it does not complete the set is 0.9. The probability that neither of the next two cards completes the set is therefore . In a similar way, the probability that none of the next cards completes the set is . This must be less than 0.01, if we are to have a probability of success > 0.99.
Can you complete the question now?