Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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

share|improve this answer
    
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 $$P_\text{xor}= 0.2\cdot 0.6 + 0.8\cdot 0.4 = 0.44 $$

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.