Thread: probability distribution

When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% had impurity B. (Obviously, some had both.) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well.

I know y = 0 is 20%.

But how is y = 1 70% and y = 2 10%?

Originally Posted by Chizum

When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% had impurity B. (Obviously, some had both.) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well.
I know y = 0 is 20%.
But how is y = 1 70% and y = 2 10%?

See the Venn diagram.
Y=1 means exactly one. A and not B or B and not A: .3+.4=.7
Y=2 means both A & B: .1

Do you find A & B by .7 + .2 + ? = 1?

Or do you calculate it somehow? It seems like, to calculate it, A and B would have to be dependent and you'd have to find P(B|A).

Originally Posted by Chizum

Do you find A & B by .7 + .2 + ? = 1?
Or do you calculate it somehow? It seems like, to calculate it, A and B would have to be dependent and you'd have to find P(B|A).

I think that you need a sit-down (face to face) session with your instructor.

Originally Posted by Plato

I think that you need a sit-down (face to face) session with your instructor.

Granted, but let me ask:

Is it possible to find P(A and B) without having first found the individual probabilities of A and B (A = 30% and B = 40%)?

I just want a yes or no. Then I'll leave.