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Question 1: Probability of License Plates (20 marks) In some states, license plates have 5 characters: three letters followed by 2 numbers. 1. How many distinct such plates are possible? (5 marks) Answer - Let L represent letters A to Z (as there are 26 letters in the alphabet and any are allowed to appear more than once there are 26 possible options in L alone). Let D represents digits 0 to 9 including 0 there are 10 digits that can appear for D alone. A License plate is created in this format - L,LL,D,D. EX. {D,E,H,3,8} [ *[*[*D*D = 26*26*26*10*10 = 1,757,600 2. If all sequences of five characters are equally likely, what is the probability that the license plate for a new car will contain no duplicate letters or numbers? (5 marks) Answer- A License plate is created in this format - L,L,L,D,D. EX. {D,E, H,3,8} In order to check the probability that a new car has no duplicate letters or numbers (L and D). Must be reduced by one the total possible combination for each (L and D) indicating that that variable can no longer be used again. P = [*[*[*D*D = (1/26)*(1/25)*(1/24)*(1/10)*(1/9) = 0.00000071225 OR (26*25*24*10*9)/ (26*$3)*(10*2) = 0798816... 3. What is the probability that a randomly selected license plate contains both A and 1? (5 marks) Answer- A License plate is created in this format - L,L,L,D,D. EX. {D,E, H,3,8} The Probability of a license plate containing both A (1/26) and 1(1/10) is: P=(1/26)*(1/10) = 0.00384615 (1/26)*(1/26)*(1/26)*(1/10)*(1/10) = 5.689576695493855e-7 OR The number of possible combination where plates contain A and 1 is :6(AAA 11), (AAB 11), (ABB 11), (AAA 10),(AAB 10), (ABB 10). Next number has to be divided by total number of possible outcomes Therefore the probability is: 1757600 4. What is the probability that a randomly selected license plate contains both A and 1 given we know that the license plate contains no duplicate letters or numbers? (5 marks) Answer- P=(1/26)*(1/10) = 0.00384615 (1/26)*(1/25)*(1/24)*(1/10)*(1/9) =0.00000071225 (0.00384615 *0.00000071225) = 2.7394203375e-9 The number of plates where A and 1 is present and there are no duplications is 1 (ABC 10). Therefore 1757600