Set Question : Probality

This topic has expert replies
Source: — Problem Solving |

User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pm
Location: Silicon valley, California
Thanked: 30 times
Followed by:1 members

Re: Set Question : Probality

by jayhawk2001 » Thu Jun 14, 2007 1:51 pm
simplythebest wrote:A random number N is chosen from 1 to 96 inclusive, what is the probability that the number N(N+1)(N+2) is divisible by 8?
5/8 ?

If n is even, then n*(n+2) will yield a product which is a multiple of 8. So,
1/2 the numbers will satisfy this.

If n is odd, the condition will be satisfied when n+1 is a multiple of 8.
There is 1 such number for every 4 consecutive odd numbers (7,
15, 23 etc.).

So, total = 1/2 + 1/2*1/4 (where 1/2 is probability of odd and
1/4 is probability of n+1 being multiple of 8 in the list of odd integers).

So, 1/2 + 1/8 = 5/8

Master | Next Rank: 500 Posts
Posts: 174
Joined: Mon Mar 26, 2007 5:51 am
Thanked: 1 times

by 800GMAT » Thu Jun 14, 2007 8:31 pm
agree with 5/8

all even numbers will satisfy the given condition
i.e. 96/2 = 48
when n is odd, multiples of 8 will satisfy the given condition i.e. 12 multiples

(48+12)/96 = 5/8

Master | Next Rank: 500 Posts
Posts: 174
Joined: Mon Mar 26, 2007 5:51 am
Thanked: 1 times

by 800GMAT » Sun Jun 17, 2007 10:23 pm
what is the OA...........

Junior | Next Rank: 30 Posts
Posts: 15
Joined: Fri Dec 15, 2006 4:25 am

by AMalik » Mon Jun 18, 2007 1:46 am
Hi,

From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*8)(7*8*9)(8*9*10)

So till 96, there will be 12 * 5 such sets = 60 sets

so probability will be 60/96=5/8

Junior | Next Rank: 30 Posts
Posts: 22
Joined: Mon Sep 18, 2006 10:17 am

by simplythebest » Mon Jun 18, 2007 2:26 am
Perfect...