In a sales campaign, a company gives each customer who buys its magazine a picture card. There are 10 different pictures and any customer who collects a complete set of all 10 picture cards gets a free gift. On any occasion when a customer buys a magazine, the card received is equally likely to be any of the 10 picture cards.

(i) Find the probability that the first 4 cards the customer receives all carry different pictures.

(ii) Find the probability that the first 4 cards received result in the customer having exactly 3 different pictures.

(iii) Two of the 10 pictures are X and Y. Find the probability that the first 4 cards received result in the customer having a picture of X or of Y (or both).

(iv) At a certain stage the customer has collected 9 of the 10 pictures. Find the least value of n such that the probability of at most n more cards are needed to complete the set is at least 0.99.

Community Answer

(i) The probability that the first 4 cards the customer receives all carry different pictures is  1st card can be anyone so 10 favorable outcomes 2nd card should be picked from the remaining 9 cards (different from 1st card) similarly, 3rd and 4th cards have 8 and 7 favorable outcomes respectively. P=(10 xx9xx8xx7)/(10^(4)) P=0.504 ---------------------------------------------------------------------------------------------------------------------------------------------------- (ii) The probability that the first 4 cards received result in the customer having exactly 3 different pictures is Any one of the 3 cards selected can be repeated, Out of 4, any two can be similar, i.e., ^(4)C_(2)=(4!)/(2!xx2!)=6 so, the total number of favorable outcomes is 6xx10 xx9xx8  P=(6xx10 xx9xx8)/(10^(4)) P=(4320)/(10^(4)) P=0.432 -------------------------------------------------------------------------------------------------------------------------------------------------------- (iii) Total number of combinations are ... See the full answer