Question 1204804
<pre>
You should make a tree diagram like this:

{{{drawing(8400/11,400,-2,40,-11,11,

locate(0,1,matrix(1,6,Box,"1:",8,Std,1,Non)),
locate(0,0,matrix(1,6,Box,"2:",3,Std,1,Non)),

locate(11,5,matrix(1,6,Box,"1:",7,Std,1,Non)),
locate(11,4,matrix(1,6,Box,"2:",4,Std,1,Non)),

locate(11,-3,matrix(1,6,Box,"1:",8,Std,0,Non)),
locate(11,-4,matrix(1,6,Box,"2:",3,Std,2,Non)),

locate(22.3,5.5,matrix(1,3,Gets,3,Std)), locate(22.3,3,matrix(1,5,Gets,2,Std,1,Non)),

line(8,0,11,3),line(8,0,11,-4),


  
locate(8.4,2.5,"8/9"),locate(8.3,-2,"1/9"),locate(20,5.7,"3/10"),

line(19,4,22,5), line(19,4,22,2.5,"7/10"),
line(19,-4,22,-2),line(19,-4,22,-4),line(19,-4,22,-6),

locate(22.3,-1.5,matrix(1,3,Gets,3,Std)), locate(20,3,"7/10"),  

locate(19.3,-2,"1/10"),locate(20,-3.2,"6/10"), 
locate(19.3,-5,"3/10"),
locate(22.3,-3.5,matrix(1,5,Gets,2,Std,1,Non)),
locate(22.3,-5.5,matrix(1,5,Gets,1,Std,2,Non)),

line(27,5,30,5),line(29.5,2.5,30,2.5),line(27,-2,30,-2),line(29.5,-4,30,-4),
line(29.7,-6,30,-6),
locate(30.3,6,prob=expr(8/9)*expr(3/10)=4/15), locate(30.3,3.5,prob=expr(8/9)*expr(7/10)=28/45),
locate(30.3,-1,prob=expr(1/9)*expr(1/10)=1/90),
locate(30.3,-3,prob=expr(1/9)*expr(6/10)=1/15),
locate(30.3,-5,prob=expr(1/9)*expr(3/10)=1/30)
  

)}}}

Notice that the sum of the probabilities on the far right is 1.
{{{4/15+28/45+1/90+1/15+1/30}}}{{{""=""}}}{{{1}}}.
There are 2 cases where he gets 3 standards, the top one with probability
4/15 and the third one down with probability 1/90.

So the probability that all three parts taken will be standard is

4/15 + 1/90 = 5/18.  I did not explain here how I got some of those
probabilities.  If you have any questions, ask me in the thank-you note
space below.

Edwin</pre>