Hi KevoK,Great question — there IS a slightly slicker way to organize this. Instead of listing all
4 winning scenarios separately, try conditioning on one VP.
Pick the easiest VP to work with. Here, Baker is nice because P(B) =
0.5 and P(B') =
0.5.
Split into two worlds:
World 1: Baker approves (probability 0.5)Johnson now only needs
1 more approval from Adams or Corfu. It's easier to find the
complement — neither approves:
0.3 ×
0.6 =
0.18. So P(at least
1 of them approves) =
1 -
0.18 =
0.82.
World 2: Baker doesn't approve (probability 0.5)Johnson now needs BOTH Adams AND Corfu:
0.7 ×
0.4 =
0.28.
Combine:0.5 ×
0.82 +
0.5 ×
0.28 =
0.41 +
0.14 =
0.55Answer: DWhy this is faster: instead of computing
4 separate three-way products and adding them up, you only need
2 quick sub-calculations. The trick is to
condition on one person, which reduces the 'at least
2 out of
3' problem into simpler 'at least
1 out of
2' and 'both out of
2' problems.
General principle: When a probability problem feels calculation-heavy, try splitting on one variable to break it into simpler sub-problems. This
'divide and conquer' approach often cuts the work roughly in half.