Question 1075831
A container holds 15 pennies, 8 nickels, and 10 dimes. 
You will randomly select two coins without replacement. 
-->Fill in the probabilities on a tree diagram.
<pre>
Here's the probability tree:

{{{drawing(400,400,-1,25,-11,11,

line(0,0,4,6.5),
line(0,0,4,0),
line(0,0,4,-6.5),
line(5,7,9,9),
line(5,7,9,7),
line(5,7,9,5),
line(5,0,9,2),
line(5,0,9,0),
line(5,0,9,-2),
line(5,-7,9,-5),
line(5,-7,9,-7),
line(5,-7,9,-9),
locate(4.5,7.4,P),
locate(4.5,.4,N),
locate(4.5,-6.6,D),
locate(9.3,9.5,P),locate(10,9.5,matrix(1,3,"<--outcome","#1",prob="35/176")),
locate(9.3,7.5,N),locate(10,7.5,matrix(1,3,"<--outcome","#2",prob="5/44")),
locate(9.3,5.5,D),locate(10,5.5,matrix(1,3,"<--outcome","#3",prob="25/176")),
locate(9.3,2.5,P),locate(10,2.5,matrix(1,3,"<--outcome","#4",prob="5/44")),
locate(9.3,0.5,N),locate(10,0.5,matrix(1,3,"<--outcome","#5",prob="7/132")),
locate(9.3,-1.7,D),locate(10,-1.7,matrix(1,3,"<--outcome","#6",prob="5/66")),
locate(9.3,-8.5,D),locate(10,-8.5,matrix(1,3,"<--outcome","#9",prob="15/176")),
locate(9.3,-6.5,N),locate(10,-6.5,matrix(1,3,"<--outcome","#8",prob="5/66")),
locate(9.3,-4.5,P),locate(10,-4.5,matrix(1,3,"<--outcome","#7",prob="25/176")),
locate(.4-.7,4,"15/33"),
locate(.4-.7,-3.5,"10/33"),
locate(2-.7,1,"8/33"),
locate(6.2-.7,9.2,"14/32"),
locate(6.2-.7,2.2,"15/32"),
locate(6.2-.7,-4.7,"15/32"),
locate(7.3-.7,7.9,"8/32"),
locate(7.3-.7,.9,"7/32"),
locate(7.3-.7,-6.1,"8/32"),
locate(7.3-.7,6.7,"10/32"),
locate(7.3-.7,-.4,"10/32"),
locate(7.3-.7,-7.2,"9/32"))}}}

There are 9 outcomes.  The probability of each outcome
is gotten by multiplying the probabilities along the two
lines from the beginning of the tree to the outcome.

For outcome #1, prob = (15/33)(14/32) which reduces to 35/176
For outcome #2, prob = (15/33)(8/32) which reduces to 5/44
For outcome #3, prob = (15/33)(10/32) which reduces to 25/176
For outcome #4, prob = (8/33)(15/32) which reduces to 5/44
For outcome #5, prob = (8/33)(7/32) which reduces to 7/132
For outcome #6, prob = (8/33)(10/32) which reduces to 5/66
For outcome #7, prob = (10/33)(15/32) which reduces to 25/176
For outcome #8, prob = (10/33)(8/32) which reduces to 5/66
For outcome #9, prob = (10/33)(9/32) which reduces to 15/176
</pre>

      --> How many ways can you select the coins?<pre>There are 9 outcomes</pre>
      --> How many ways can you select exactly 1 nickel?<pre>That's either outcome #2, #4, #6, or #8, that's 4 ways</pre>
      --> What is the probability that you select 2 pennies?<pre>
That's outcome #1, probability = 35/176</pre>
      --> What is the probability that you select a dime and then a penny?<pre>
That's outcome #7, probability = 25/176

You got the hardest two right, and missed the two easy ones. lol

Edwin</pre>