Question 347965
You have 5 keys on a key ring, 2 of the keys unlock the door. Find the 
probability of opening the door on the first or second try. Draw a tree diagram 
to illustrate.
<pre>
{{{drawing(800,400,-1,10,-3,3,

line(0,0,1,1), line(0,0,1,-1), line(2.1,-1,3.1,-2), line(2.1,-1,3.1,0),
locate(1.1,1.1,"RIGHT_KEY------------------------->"), locate(1.1,-1+.1,WRONG_KEY),
locate(.3,1,2/5),locate(.5,.1,3/5), locate(3.2,.15,"RIGHT_KEY------>"),
locate(3.2,-1.9,"WRONG_KEY------>"),locate(2.1+.3,-1+1,2/4), locate(2.1+.5,-1+.1,2/4), locate(5,1.5,OUTCOME), red(locate(5,1.1,success)),
red(locate(5,.15,success)), locate(5,-1.9,failure), locate(7.5,1.5,PROBABILITY),
red(locate(8,1.1,2/5)), red(locate(7,.2,(3/5)(2/4)=(3/5)*(1/2)=3/10)),
locate(7,-1.7,(3/5)(2/4)=(3/5)*(1/2)=3/10) 

 )}}} 

Answer:  We add the two probabilities of successes

2/5 + 3/10 in red above.

                            2/5 + 3/10 = 4/10 + 3/10 = 7/10

Or we could subtract the probability of two wrong keys 3/10 from 1
and get 1 - 3/10 = 7/10.

Explanation:  The upper left branch represents picking a right key
the first time with a probability of 2 right keys out of 5, or 2/5.
That led to a success (opening the door).  So we write 2/5 on the
branch that leads to that, and we do not branch from because we
open the door. 

The lower left branch represents picking a wrong key the first time 
with a probability of 3 wrong keys out of 5, or 3/5. So we write
3/5 on the lower left branch.  So we must branch again from there.
There are two possibilities for the second key we try.  We have
eliminated 1 wrong key, so we only have 4 keys to try, 2 are wrong
keys, and 2 are right keys.  So it's a probability of 2 right keys 
out of 4 for picking the right key the second time and also a 2 
wrong keys out of 4 for picking the wrong key the second time. That's
why I wrote 2/4 instead of reducing it to 1/2.

Edwin</pre>