PMV wrote:Hi guys! I'm new in theory of probability and I'm having a hard time solving a probability problem. I'll be really grateful if you solve it and if possible explain to me the solution.

So here's the problem:

Test for impurities commonly found in drinking water from private wells showed that 30% of all wells in a particular country have impurity A. If a random sample of five wells is selected from the large number of wells in the country, what is the probability that:

a) Exactly three will have impurity A?

b) At least three?

c) Fewer than three?

Using Binomial Distribution

= n!/(n-x)! x! (p^x) (q^n-x)

where n = 5, p = 30% = 0.3, q = 1-p = 1-0.3 = 0.7

a) P(Exactly 3) = P(x=3) = 5!/(5-3)!(3)! (0.3^3) (0.7^5-3)

= 10(0.027)(0.49)

= 0.1323

b) Prob(At least 3) = P(x >= 3) = P(x=3) + p(x=4) + p(x=5)

= (5!/(5-3)!(3)! (0.3^3) (0.7^5-3)) + (5!/(5-4)! (0.3^4) (0.7^5-4) + (5!/(5-5)! (0.3^5) (0.7^5-5)

= 10(0.027)(0.49) + 5(0.0081)(0.7) + 1(0.00243)(1)

= 0.1323 + 0.02835 + 0.00243

= 0.1631

c) Prob(Fewer than three) = P(x = lessthan 3) = 1 - P(X greater than or equals to 3)

= 1 - (5!/(5-3)!(3)! (0.3^3) (0.7^5-3)) + (5!/(5-4)! (0.3^4) (0.7^5-4) + (5!/(5-5)! (0.3^5) (0.7^5-5)

= 1 - 0.1631

= 0.8369