BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability that a # is formed

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 682
Joined: Fri Jan 16, 2009 2:40 am
Thanked: 32 times
Followed by:1 members

Probability that a # is formed

by Vemuri » Thu Mar 05, 2009 5:49 am
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
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 80
Joined: Mon Feb 02, 2009 6:36 am
Thanked: 10 times

Re: Probability that a # is formed

by billzhao » Thu Mar 05, 2009 7:15 am
Vemuri wrote: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
Yiliang
Top
User avatar
Legendary Member
Posts: 682
Joined: Fri Jan 16, 2009 2:40 am
Thanked: 32 times
Followed by:1 members

Re: Probability that a # is formed

by Vemuri » Thu Mar 05, 2009 7:42 pm
The answer is not B
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Thu Mar 05, 2009 8:07 pm
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
Top
Master | Next Rank: 500 Posts
Posts: 148
Joined: Wed Dec 10, 2008 5:13 pm
Location: SF, CA
Thanked: 12 times

by lilu » Thu Mar 05, 2009 10:37 pm
cramya wrote: Total = 144+72 = 216
Cramya, can you elaborate more on how you got these numbers?
Top
Senior | Next Rank: 100 Posts
Posts: 88
Joined: Tue Mar 03, 2009 8:10 am
Thanked: 3 times

by quocbao » Thu Mar 05, 2009 10:51 pm
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 :(
Top
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Fri Mar 06, 2009 2:02 am
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-
Top
Post new topic Post Reply
• Page 1 of 1