CE maths problem

Since only E, I, S, H appears in both words, we only consider P(E), P(I), P(S), P(H)
where P(X) respresents the probability that draw X in two words at the same time
P(E)=1/11*1/7=1/77
P(I)=1/11*1/7=1/77
P(S)=1/11*1/7=1/77
P(H)=1/11*1/7=1/77
So the probability that the two letters drawn are the same
=1/77*4
=4/77

(1 / 11 + 1 / 7) × 4 = 4 / 77
MATHEMATICS有11個英文字母,
ENGLISH就有7個,佢地兩個詞語有重複過既字母有四個,
估此就有"4"既存在