Question

A,b,c are three marksmen standing in a triangle.They have an accuracy rate of 99%,87%,1%).They all have one gun with one bullet each.At 12.00 pm,they shoot among themselves(all the bullets were fired at the same time).What is the chance that only one survives at the end

Answer 1

Answer:

Step-by-step explanation:

P(A)=99/100

P(B)=87/100

P(C)=1/100

∆ABC

A shoots at B and B shoots at C and C shoots at A

So P(one survivor)=99/100+87/100+1/100

=87/100