Test day tomorrow...help!!

[Sandman]

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.

[tendays2go]

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

[cramya]

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!!!