Probability that a # is formed

[Vemuri]

A 5 digit number is formed using the digits 0,1,2,3,4&5 at random but without repetition. The probability that the number so formed is divisible by 5 is:

A. 1/5
B. 2/5
C. 4/5
D. 3/10
E. 9/25

[billzhao]

A 5 digit number is formed using the digits 0,1,2,3,4&5 at random but without repetition. The probability that the number so formed is divisible by 5 is:

A. 1/5
B. 2/5
C. 4/5
D. 3/10
E. 9/25

Answer is (B)

The total number of 5-digit integers is: 5*5! (The first digit cannot be zero)
The number of integers which are divisible by 5 is: 2*5! (integers that end with 0 or 5 are divisible by 5).

Thus probability=2*5!/(5*5!)=2/5

[Vemuri]

The answer is not B

[cramya]

God please let probability be my forte since its never been....

Is it E 9/25

The 5 digit number is divisble by 5 if 5 digit number ends as 05,10,15,20,25,30,35,40,45,50.

For 6 of these combinations there will be 24 possible ways and for 4 of them that dont have 0 in either of the last 2 digits there will be 18 possible ways.


Total = 144+72 = 216

216/600 = 9/25

Regards,
CR

[lilu]

Total = 144+72 = 216

Cramya, can you elaborate more on how you got these numbers?

[quocbao]

Hmm, i got B too

let ABCDE is the number, if the number is divisible by 5 then E must be 0 or 5.

We have total 5 numbers, 2 number is possible then it is 2/5

Can someone explain why I'm wrong

[bluementor]

Assumption: The 5-digit number will not begin with 0.

The 5 digit number can be formed in the following number of ways:

5x5x4x3x2

The first(left most) digit can be either 1,2,3,4 or 5.
The 2nd digit can be either of of the remaining 5 digits.
The 3rd digit can be either of of the remaining 4 digits.
etc… (this is the peagon hole concept for counting)


A number divisible by 5 can be formed with the last digit ending with either 5 or 0.

Case 1) Number ending with digit 0:

If the last digit is 0, then the first 4 digits can be arranged in the following number of ways:

5x4x3x2

The first(left most) digit can be either 1,2,3,4 or 5.
The 2nd digit can be either of of the remaining 4 digits.
The 3rd digit can be either of of the remaining 3 digits.
The 4th digit can be either of of the remaining 2 digits.

Case 2)Number ending with digit 5:

If the last digit is 5, then the first 4 digits can be arranged in the following number of ways:

4x4x3x2

The first(left most) digit can be either 1,2,3, or 4 (0 is not allowed, and 5 is already in use)
The 2nd digit can be either of of the remaining 4 digits.
The 3rd digit can be either of of the remaining 3 digits.
The 4th digit can be either of of the remaining 2 digits.


Therefore, the probability of having a 5-digit number divisible by 5 is:

{(5x4x3x2) + (4x4x3x2)} / (5x5x4x3x2)

=(5 + 4) / (5x5)
=9/25

Choose E.

-BM-