2 ms·
I think about it this way p(th) = p(t) p(h) p(ht) = p(h) p(t) Hence p(th) = p(ht) regardless of coin imbalance as long as both events actually will happen. Q
by toxik 3y ago
I think about it this way
p(th) = p(t) p(h)
p(ht) = p(h) p(t)
Hence p(th) = p(ht) regardless of coin imbalance as long as both events actually will happen. QED.
- aqme28 3y agoYeah this is simpler. You're just throwing away every pair that isn't a TH or HT