# Probability using percentages question

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 so I'm trying to expand this out to four numbers... A = 10% B = 20% C = 30% D = 40% P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A * B) - P(A * C) - P(A * D) - P(B * C) - P(B * D) - P(C * D) + P(A * B * C * D) .10 + .20 + .30 + .40 = 1 P(A) + P(B) + P(C) + P(D) - .02 - .03 - .04 - .06 - .08 - .12 P(A * B) - P(A * C) - P(A * D) - P(B * C) - P(B * D) - P(C * D) + .0024 = .06724 67.24% – Mike Apr 10 '12 at 22:05
Your formula is incorrect; you are missing terms such as $P(A\cap B\cap C)$ which will occur with a positive sign, and $P(A\cap B\cap C \cap D)$ will be subtracted, not added the way you have it. More easily, \begin{align*}P(A\cup B \cup C \cup D)&=1-P(A^cB^cC^cD^c)\\&=1-(1-P(A))(1-P(B))(1-P(C))(1-P(D)).\end{align*} – Dilip Sarwate Apr 11 '12 at 1:04
@DilipSarwate so basically all I'm doing with 10%/20%/30%/40% is finding the probability of it NOT happening... .90 * .80 * .70 * .60 = .3024 and then 1 - .3024 = .6976 = 69.76% chance of any one of the 4 hapening? – Mike Apr 12 '12 at 1:33

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 $P(A\text{ or }B)$.

-
ok so if I wanted to expand this out to 4 "subjects" and they are all independent of each other would it then be as follows: P(A or B or C or D)=P(A)+P(B)+P(C)+P(D) − P(A and B and C and D) with P(A and B and C and D) = P(A)⋅P(B)⋅P(C)⋅P(D) ? Thanks! – Mike Apr 10 '12 at 17:28
@Mike No, your formulas are incorrect for the case of 4 events. Look at the inclusion-exclusion principle link given by Zev Chonoles. – Dilip Sarwate Apr 10 '12 at 17:49
@DilipSarwate so... this is what i've got after reading, is this correct? P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A * B) - P(A * C) - P(A * D) - P(B * C) - P(B * D) - P(C * D) + P(A * B * C * D) ugh this comments box screws up the formatting but yea... – Mike Apr 10 '12 at 21:34

"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 $$P_\text{xor}= 0.2\cdot 0.6 + 0.8\cdot 0.4 = 0.44$$

-
There is an implicit assumption of independence here that should be mentioned explicitly. – Dilip Sarwate Apr 10 '12 at 17:48