How do you determine the probability that four marksmen will all hit a target if each marksman usually hits the target 80 percent of the time?

Answer 1

You would multiply the probabilities. The probability the first marksmen hits the target x the probability the second marksmen hits the target x the probability the third marksmen hits the target x the probability the fourth marksmen hits the target. So you take .80 x .80 x .80 x .80 = .4096 or about 41%. So if they all fire at the target with each having an 80% probability of hitting, there will be about a 41% chance they will all hit. If you actually think about this question, though, you would be wise to hesitate about the answer. What causes marksmen to miss? In general it would be many things, often acting in combination: trembling of the hands, distractions, shifts of the wind, variations in the ammunition being fired, and so on. If all four marksmen are shooting simultaneously, or nearly so, then some of these causes will be acting in the same way on all four. These would include the wind and the environmental distractions, for instance. It's therefore conceivable that when one marksman misses, so will all the others, and (usually) when one marksman hits the target, the others will be able to as well. In this situation the probability that four marksmen will all hit the target will be close to 80%, not 41%. The question itself couches some ambiguities. For instance, at a tournament of 100 marksmen, the probability that some four will all hit their target is likely close to 100%. (Problems like this in interpreting the intended meaning of probability questions go all the way back to the very first book on probability by Christian Huygens in 1657. There was argument among very good mathematicians for a long time about the answer to one of his problems because it had three distinct interpretations.)