Question 1055395
So the probability for both Wally and Sally for number of heads would be calculated,
{{{P(X)=C(8,X)*(0.5)^X*0.5^(8-X)=C(8,X)*0.5^8=C(8,X)/256}}}
So then find the joint probability when Sally has 5 more than heads than Wally
Sally-5, Wally-0: {{{P=(1/256)(56/256)=56/256^2}}}
Sally-6, Wally-1: {{{P=(8/256)(28/256)=224/256^2}}}
Sally-7, Wally-2: {{{P=(28/256)(8/256)=224/256^2}}}
Sally-8, Wally-3: {{{P=(56/256)(1/256)=56/256^2}}}
So summing those together,
{{{P=(56+224+224+56)/256^2=560/256^2=35/4096}}}