Question

Consider all seven-digit numbers that can be created from the digits 0-9 where the first and last digits must be odd and no digit can
repeat. What is the probability of choosing a random number that starts with 5 from this group? Enter a fraction or round your answer to
4 decimal places, if necessary.

Answer 1

Answer:

1/4 or 0.25

Step-by-step explanation:

The total possibilities of any 7 digit number using 0-9 is :

9×10×10×10×10×10×10=9000000

To work out the total possibilities in this question :

We look at the conditions :

The first digit can only be 5 numbers :

1 , 3 , 5 , 7 , 9

Now we subtract 5 from 9 :

9-5 = 4

Since no repeats for 2 , 3 , 4, 5, 6:

9 , 8 , 7 , 6 , 5,  

5 possibilities for the last digit :

Total possibilities for this code :

4 × 9 × 8 × 7 × 6 × 5 × 5 = 302400

If it begins with 5 that is only 1 possibility for the first digit

1 × 9 × 8 × 7 × 6 × 5 × 5 = 75600

Now we make a fraction :

75600÷302400

Dividing top and bottom by 75600 gives you 1/4 or 0.25

Hope this helped and have a good day

Answer 2

Answer:

Step-by-step explanation:

Comment

The first digit and the last digit are both odd. That tells you that so far what you have is one of 5 digits for the first digit and and one of 4 for the last digit. 4 because you can't repeat the first digit.

5, , , , , ,4

2 digits are gone 8 remain.

5* 8 * 7* 6* 5* 4* 4 = 134400

Part 2

Only one number can go at the beginning, and that is a 5. Everything else remains the same.

1 * 8 * 7 * 6 *5 * 4 * 4 = 26880

P(picking a number beginning with a 5 is  25880 /13440) = 0.2