I think the answer is 48/125.

Here's how I did it:

For the denominator,
Since repitition of digits is allowed, there are five ways to pick the first digit, five ways for the second digit,..., basically five ways to pick each of the five digits (1,2,3,4,5).
So, the total number of five-digit numbers you can have is: 5*5*5*5*5 = 3,125.

For the numerator,
Since the question is asking for the probability of picking a five-digit number that has only one digit repeated with two occurences, you can think of the number as, for example, a five digit number with two 1's:
11234 =>These digits can be arranged in 5! ways to form 5! = 120 numbers.
Similarly, it is also possible to have the following number with two 1's:
11235 (we didn't use the 5 above) => These digits as well, can be arranged in 5! = 120 ways (same as above).
So, in total you'll have 240 different five-digit numbers that have two 1's. Therefore, you'll have 240*5 (numbers with two 1's, or two 2's,..., or two 5's), for a total of 1,200 different five-digit numbers that have only one digit repeated two times.

So, the final probability is 1,200/3,125 = 48/125 (Option B).

-Ash

I too find the answer to be 48/125.

11234 =>These digits can be arranged in 5! ways to form 5! = 120 numbers.

no ashish.These digits cannot be arranged in 5! ways.They can be arranged only in 5!/2! ways

12341 is same as 12341!

Rule : The number of ways n digits can be arranged with r similar digits is n!/r!

I go with None of these. The answer i got is 24/ 625.

Here is the logic i used -
Total number of ways - 5^5 = 3125

As for the number of ways to arrange the 5 digits in the 5-digit number with only one digit repeated with two occurences -

11--- the rest of the numbers can be filled in 4*3*2=24 ways.
for 22--- same as above - 24 ways.

Similarly for all the 5 numbers it is 24*5= 120 ways.

So the probabilility is 120/3125 = 24/625.
So none of these.
Let me know if this logic is incorrect.

Guys by my logic its 48/625 just now checked it.

my logic,

case 1:

11 x y z - there are 4 options(2,3,4,5) for this 3 numbers x,y,z

there are 4c3 combinations of (2,3,4,5) = 4 combinations

for (1,1,2,3,4) we get 5!/2! arrangements = 60

so 60 * 4(different combinations) = 240 for (1,1,x,y,z)

for all 240 * 5 = 1200

1200/5*5*5*5*5* = 48/625

total poss. = 3125

11_ _ _ = 5!/2! * 4c3 = 240 (4c3 comes from the fact that you need to choose three from 2,3,4,5)

22_ _ _ = 5!/2! * 4c3 = 240

33_ _ _ = 5!/2! * 4c3 = 240

44_ _ _ = 5!/2! * 4c3 = 240

55_ _ _ = 5!/2! * 4c3 = 240

240*5 = 1200

Therefore; 1200/3125 = 48/125