If four digit numbers greater than 5000 are randomly formed from the digits 0, 1, 3, 5 and 7,what is the probability of forming number divisible
by 5 when
(i) the digits are repeated ?
(ii) the repetition of digits is not allowed ?

 Case I When digits are repeated.
Total four digit numbers which are greater than 5000 from the digits 0, 1, 3, 5,7 i.e. in thousand place, the 
number will be 5 or 7 and rest of three places any five numbers will come

$5 \underset{\_}{5}\times \underset{\_}{5}\times \underset{\_}{5}-1=124$

(here, one is subtracted because one case is possible in which all the three place are occupied by the number zero.)
$\text{ 7 }\underset{\_}{5}\times \underset{\_}{5}\times \underset{\_}{5}=125$
$\therefore$ Total four digit numbers, n (S) = 124 + 125 =249
Total four digit number which are divisible by 5.
i. e , In unit place number will be 0 or 5 and thousand
place the number will be 5 to 7. And rest of the two places any five numbers will be come. 

Number of favorable cases, n(E) =24+25+25+25=99
Required probability $=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{99}{249}=\frac{33}{83}$ 

Case II When digit are not repeated.
Total four digit number which are grater than 5000
&om the digits 0, l,5,7,5 i.e, In thousands place, the
number will be 5 or 7. Since, the digit are not repeated,
then rest of the place may be filled by 4, 3, 2.
$5\underset{\_}{4}\times \underset{\_}{3}\times \underset{\_}{2}=24$

$7\underset{\_}{4}\times \underset{\_}{3}\times \underset{\_}{2}=24$
Total four digit number n(S) = 24 + 24 = 48
Total four digit numbers which are divisible by 5

Favorable number of cases n(E)=6 + 6 + 6 = 18
.'. Required probability $=\frac{18}{48}=\frac{3}{8}$