Question

exercise 1.3. a telephone number consists of ten digits, of which the first digit is one of 1;2;:::;9 and the others can be 0;1;2;:::;9. what is the probability that 0/ appears at most once in a telephone number, if all the digits are chosen completely at random

Answer 1

Answer:

The probability that "0" appears at most once in a telephone number is 0.82

Step-by-step explanation:

Telephone number = 10 digits

First digit = (1,2......9)------> there are 9 possible digits

digit 2 - digit 10 = (0,1,2......9) -----> there are 10 possible digits

Let's find the total amount of telefone numbers with those characteristics:

possible first digit X possible second digit X...............X possible tenth digit

9X10X10X10X10X10X10X10X10X10 = $9X10^{9} = 9.000.000.000$

Let's find the amount of telephone numbers that "0" appears at most once (it means 0 will occur either once or will not occur):

If "0" appears once, the rest of the numbers should be 1.2.....9 (9 possible digits)

9X10X9X9X9X9X9X9X9X9 =$10X9^{9} = 3.874.204.809$

If "0" doesn't appear, the ten digits should be 1,2....9

9X9X9X9X9X9X9X9X9X9 = $9^{10} =3.486.784.401$

Finally, find the probability

P=(favorable outcomes/total outcomes)

P=$\frac{3.874.204.890 + 3.486.784.401}{9.000.000.000}$ = 0.82