Let E1: choosing first (two headed) coin, E2 : choosing second (biased) coin, E3 : choosing third coin. Also, let A : the coin showing heads.
∴P(E1)=P(E2)=P(E3)=13,P(A|E1)=1,P(A|E2)=75100,P(A|E3)=60100.
By Bayes' Theorem, P(E1|a)=P(A|E1)P(E1)P(A|E1)P(E1)+P(A|E2)(P(E2)+P(A|E3)P(E3)
=13×113×1+13×34+13×35=2047.
OR
Total number of ways of selecting two numbers form six positive integers = 6C2=6×52=15 .
Let X denotes the larger of the Two numbers selected, So, values of X:2,3,4,5,6.
So, P(X=2)=115,P(X=3)=215,P(X=4)=315,P(X=5)=415,P(X=6)=515
∴ distribution can be written as:
X23456P(X)115215315415515
Therefore, mean of the distribution =∑XP(X)=2×115+3×215+4×315+5×415+6×515=7015=143