# probability distribution

• Sep 17th 2009, 02:32 PM
Chizum
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%?
• Sep 17th 2009, 03:04 PM
Plato
Quote:

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
• Sep 17th 2009, 03:52 PM
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).
• Sep 17th 2009, 04:10 PM
Plato
Quote:

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.
• Sep 17th 2009, 04:50 PM
Chizum
Quote:

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