Question

Felecity cannot remember the correct order of the five digits in her ID number. She does remember that the ID number contains the digits 5, 0, 3, 7, 4. What is the probability that the first three digits of Felecity’s ID number will all be odd numbers?

Answer 1

Answer:

The probability is:

                   $\dfrac{1}{10}$      

Step-by-step explanation:

The ID number contains the digits: 5,0,3,7,4

The number of ID which contain the odd number in the first three places is:  

  3!×2!=3×2×1×2×1=  12

( Since, in the first place any of the three odd number could come up, in the second place any of the remaining two odd number could come up and in the third place the remaining odd number will come up.

Similarly in the fourth place any of the two non-odd number could come and at the last place the remaining non-odd number will come up)

Total number of ID numbers that are possible are:

5×4×3×2×1=120

( Since at the first place any of the number could come up, in the second place any of the remaining four number could come up, in the third place any of the remaining three numbers could come up and so on)

Hence, the probability that the first three digits of Felecity’s ID number will all be odd numbers is:

            Ratio of the numbers such that the first three numbers are all odd to the total number of ID number possible

i.e. The probability is:

$\dfrac{12}{120}=\dfrac{1}{10}$

Answer 2

Answer:

3!2!/5! or 3/5 x 2/4 x 1/3 either way it makes 6/60 = 1/10