a. We can choose every digit from 0,1,2,3,4,5,6,7,8,9. So we have 10 options ( n1=10, n2=10, n3=10, n4=10, n5=10).
Sample Space N=n1xn2xn3xn4xn5=105
b. To choose first digit we have 10 options, second digit - 9 options, third - 8 options, fourth - 7 options, five's - 6 options. So code with no repeats can be obtained in N1= 10x9x8x7x6= 30240 options. Probability that the code contains no repeats is N1/N =30240/105 =0.3024
c. Even digits are 0,2,4,6,8. So we have 5 options of the last digit. So code ends with an even number can be obtained in N2=5x104 =50000 options. Probability that the code ends with an even number is P=N2/N=5x104 /105 =0.5000
d. Odd numbers are 1,3,5,7,9. So we have 5 options for choosing each digit. So code containing only odd numbers can be obtained in N3=55=3125 options. Probability that the code contains only odd numbers is P=N3/N=3125/105 =0.0313
e. We have only one option for the last digit is 2. So code ends with 2 can be obtained in N4=1x104 options. Probability that the code ends with 2 is P=N4/N=104/105=0.1000.
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