John is drinking a glass of water with exactly 2 bacteria suspended in it, at random positions. If he drinks ½ the water, what are the probabilities that he drinks zero or one bacterium respectively?

Answer 1

Strategy: Use Pascal's Triangle (or derivatives of it) to solve discrete math problems of this type.

Tactics: Pascal's Triangle shows possible distributions of n objects randomly amongst n equal-sized containers. Thus, in conventional form :

1

1 2 1 → this line used below.

1 3 3 1 etc.

shows numbers of ways of obtaining, for the rows respectively, 1 object in 1 container, 2 objects in 2 containers, 3 objects in 3 containers, etc. For this problem, the second row applies: there is one way (hence one probability-portion) of putting both objects in the first container, two ways of assorting them variously one to each container (hence two probability-portions), and 1 way of putting them both into the second container (hence one probability-portion). Then for the first container only (the ½ of the water John drinks), only 1 of these 4 ways has that container empty (because both objects are in the second container, i.e. in the the remaining intact ½ of the water); but 2 of the 4 ways have a single object in container 1. So the respective probabilities are (1/4, 1/2).

One does note that due to vagaries of municipal water supplies, water is seldom free from bacteria, viruses, etc.; we do hope for absence of pathogens.

However, what is deemed a pathogen may change over time: Candida auris wasn't even on the screens a decade ago, but has suddenly emerged as a multidrug-resistant, lethal opportunistic pathogen. And we're encouraging multidrug-resistance by our treatment of certain sewage and wastewater; old treatment protocols aren't sufficient to prevent spread of that capability across bacterial and fungal domains. There is considerable need for new inventive science there!