Exam day after tomorrow - plz help with this one in prob..

[msgrewal81]

In MGMAT SG 4, Page 98, there are 2 approaches and I will use both but I dont know why the answer is different.

Example problem on MGMAT SG 4 PAGE 98:
There are 6 team members. 4 have to be selected. What is probability that 2 particular members, say X and Y, are both selected.

MGMAT approach 1:
Total possible outcomes = 6! / (4! x 2!) = 15 teams
Winnining outcomes where 2 particular members are on team= 4! /(2! x 2!) = 6
Therefore probability that 2 particular members on team = 6/15 = 2/5

Approach 2:
let X be selected first, Y second, "Some other" third, "Some other" fourth - to make a four member team out of 6.

Therefore probability = 1/6 x 1/5 x 4/4 x 3/3 = 1/30
But since order does not matter, actual probability = 1/30 x 4! = 4/5

OA : 2/5
What's wrong with approach 2 ???

[Ben Ku]

A problem from our Quant Strategy Guide should go in the Quant Strategy Guide folder.

In your approach, you first found the probability of selecting XYZZ (the last two are Z because it doesn't really matter).

In the second step, we want to find all the different ways of arranging XYZZ. This should be 4! / 2! (you divide by 2! because Z is repeated). So you should multiply 1/30 by 4!/2! (instead of just 4!).

Hope that helps.