From word possession we take 4 letters without
repetition, what is the probability to have oops?
How do we solve this problem?
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
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.
EDIT : refer stolyar's post, i made a mistake with the bold part...its been corrected now