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

Redeem

Test day tomorrow...help!!

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Wed Mar 05, 2008 10:21 am

Test day tomorrow...help!!

by Sandman » Wed Sep 24, 2008 4:30 am
Can someone plz help me with these probability problems. I usually end up getting them wrong.

1. A and B work on a gas station with 4 other workers. For an interview, 2 of the 6 workers will be chosen at random. what is the prob. that A and B both will be chosen?

a. 1/15
b. 1/12
c. 1/9
d. 1/6
3. 1/3

2. if 2 of the 4 expressions x+y,x-y,x+5y and 5x-y are to be chosen at random, what is the prob that the product will be in the form x^2 - (by)^2 where b is an integer.

a. 1/2
b. 1/3
c. 1/4
d. 1/5
e. 1/6


my test day is tomorrow..i gave 2 gmat prep tests and got 650 on both..hoping to get more that 650 tomorrow...pray for me guyz...plz.
Top
Source: — Problem Solving |
User avatar
Senior | Next Rank: 100 Posts
Posts: 55
Joined: Sun Sep 14, 2008 6:51 am
Location: Netherlands
Thanked: 10 times
GMAT Score:680

by tendays2go » Wed Sep 24, 2008 4:43 am
hello..if it's not pretty late then here's my reply:
(btw, i reckon both of these are GMATPrep quesn)

Q1: p(A,B) is calculated by:
choosing AorB out of 6 workers: 2/6
now, let's say we have either one of them chosen up, then the remaining ways of choosing the other is 1/5.
since both of these are independent events. thus, (2/6)*(1/5) = 1/15

Q2: here only multiplication of 'x+y' & 'x-y' gives me the required form of resulting expression x^2 - (1*y)^2 :)
so similarly, chances of choosing either of them: 2/4 and once one is in the bag, chance of choosing other is: 1/3
therefore, combining them, we get the result (just like prev quesn)
2/4 * 1/3 = 1/6 [ And, i know this one is right for sure ;) ]

All the best, Sandman
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Wed Sep 24, 2008 4:46 am
1) 1c1.1c1 / 6c2 = 1/15

2) On a smilar note the prob = 1c1.1c1/4c2 = 1/6

Alternate solution:

P(x+y) picked * p(x-y) picked + p(x-y)picked * p(x+y) picked

= 1/4*1/3+1/4*1/3
=2/12
=1/6

I am pretty sure these are the solution but kindly wait for other answers too just to be 100% sure.

All the best for your exam; Be confident and succeed!!!
Top
Post new topic Post Reply
• Page 1 of 1