Ok this is probably an easy one,
Person A hits a target 20% of the time Person B hits a target 40% of the time
What are the odds, and formula, that either one of them hits the target?
Ok this is probably an easy one,
Person A hits a target 20% of the time Person B hits a target 40% of the time
What are the odds, and formula, that either one of them hits the target?
For brevity, lets name the event that $A$ hits the target "$A$", and likewise with $B$. By the inclusion-exclusion principle, we know that $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B).$ (This is also called the "addition law".)
Also, the implicit assumption is that $A$'s success in hitting the target is independent of $B$'s, i.e. the probability $A$ hits the target doesn't change depending on whether $B$ does, or vice versa. Therefore, $P(A\text{ and }B)=P(A)\cdot P(B).$ Now you can compute the value of $P(A\text{ or }B)$.
"either one of them" is not clear.
if "either" means "one, or both" then the probability is $0.2$ that A hits and in case he doesn't ($0.8\cdot$) the other one might with a chance of $0.4$. So $P_\text{or}=0.2+0.8\cdot 0.4=0.52$
if "either" means "one, but not both", then the probability is that of A hitting, but B missing ($0.2\cdot0.6$) plut that of A missing but B hitting ($0.8\cdot0.4$). therefore $P_\text{xor}= 0.2\cdot 0.6 + 0.8\cdot 0.4 = 0.44 $