1. ## Probability Question

Hey all. I would really appreciate it if someone could help me out with this question. Thanks!

In a sales campaign, a petrol company gives each motorist who buys their petrol a card with a picture of a film star on it. There are 10 different pictures, one each of 10 different film stars, and any motorist who collects a complete set of all 10 pictures gets a free gift. On any occasion when a motorist buys petrol, the card received is equally likely to carry any one of the 10 pictures in the set. Two of the ten film stars in the set are X and Y.

Find the probability that the first four cards received results in the motorist having

(i) all different pictures
(ii) exactly three different pictures
(iii) a picture of X or of Y or of both.

At a certain stage, the motorist has collected nine of the ten pictures. Find the least value of n such that P(at most n more cards are needed to complete the set) > 0.99

2. ## Probability

Hello somethingsmells
Originally Posted by somethingsmells
Hey all. I would really appreciate it if someone could help me out with this question. Thanks!

In a sales campaign, a petrol company gives each motorist who buys their petrol a card with a picture of a film star on it. There are 10 different pictures, one each of 10 different film stars, and any motorist who collects a complete set of all 10 pictures gets a free gift. On any occasion when a motorist buys petrol, the card received is equally likely to carry any one of the 10 pictures in the set. Two of the ten film stars in the set are X and Y.

Find the probability that the first four cards received results in the motorist having

(i) all different pictures
(ii) exactly three different pictures
(iii) a picture of X or of Y or of both.

At a certain stage, the motorist has collected nine of the ten pictures. Find the least value of n such that P(at most n more cards are needed to complete the set) > 0.99
(i) Since any of the cards can be chosen from all 10, there are $10^4$ possible choices of 4 cards. Also, there are 10 x 9 x 8 x 7 ways of arranging four different cards. (Can you explain why?) So the probability that all four are different is $\frac{10 \times 9\times 8 \times 7}{10^4}$.

(ii) We have to arrange 4 cards in order, 2 of which are the same, and the remaining 2 different. There are $\frac{4 \times 3}{2} = 6$ 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 $10^4$ 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 $0.9^2$. In a similar way, the probability that none of the next $n$ cards completes the set is $0.9^n$. This must be less than 0.01, if we are to have a probability of success > 0.99.

Can you complete the question now?