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| author | Holden Rohrer <hr@hrhr.dev> | 2020-09-15 14:30:01 -0400 |
|---|---|---|
| committer | Holden Rohrer <hr@hrhr.dev> | 2020-09-15 14:30:01 -0400 |
| commit | 119687ab933828f359372927ca7d5129cb108d0b (patch) | |
| tree | 989f5f5d8dfaca5a6b65b35d2c3eb22ca87ec62e /houdre | |
| parent | 60bb75220448699a01f1c8015b5e35a8c4540e99 (diff) | |
ugh harsh grading scheme
Diffstat (limited to 'houdre')
| -rw-r--r-- | houdre/hw1.tex | 5 |
1 files changed, 5 insertions, 0 deletions
diff --git a/houdre/hw1.tex b/houdre/hw1.tex index 4baddda..c06f5fa 100644 --- a/houdre/hw1.tex +++ b/houdre/hw1.tex @@ -136,6 +136,8 @@ These are both calculated using inclusion-exclusion, the first being three independent events with probabilities $(p^2,p^2,p)$ and the second being several dependent probabilities. +% Insufficient explanation (6/10) + \q14 $$\Pr(A\cup B) = \Pr(A\setminus B)+\Pr(B\setminus A)+\Pr(A\cap B)$$ @@ -247,4 +249,7 @@ $$= 3\Pr(\hbox{matched red}) = 3{6\over(3+n)(2+n)} This means that the probability of matched black socks is the same as the probability of matched red socks: ${6\over(3+3)(2+3)} = {1\over5}$. +% This is mostly correct, but n is actually the total number of socks, +% so n=6. 2/10 + \bye |