csperber wrote:

From word possession we take 4 letters without

repetition, what is the probability to have oops?

How do we solve this problem?

hmm.without repitition??

from what i understand,we could draw poso ,sopo...etc..and still it would

mean the same, thats oops.

its not like we have to draw it in order...like first o ,second o, third p and

fourth s..

Step 1. Total # of ways

Draw 4 letters , order is not important..so we use combinations.

10C4 = 210 ways

Step 2. Favorable ways

We can draw "o's" in 1 ways...p in 1 ways...

s's in 4C1 ways
so the number of favoralbe events are 1 * 1C1 * 4C1 = 4 favorable

events which give us the required solution.

Required probability = 4/210 = 4/105 = 2/105

Someone verify please.

thanks

praetorian

EDIT : refer stolyar's post, i made a mistake with the bold part...its been corrected now