Probability Problem

[PMV]

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?

[steve kelly]

Hey guys,

I know this is a simple question/problem to you, but I would just like to know the probability of betting on a 6-digit lottery ticket where the numbers used are from 1 to 55 and are non-repeating. I know of someone who is addicted to betting and I would really want to make a point by giving him the facts and numbers. He usually bets around 20-30 tickets in one draw. Please help. Thanks =)

Steve Kelly

[Math Tutor]

There are good probability videos that could help you to solve the problems by yourselves.

[amit28it]

Two events A and B are independent. Given that[tex]P(A) = 3 P(B)[/tex], and [tex]P(A\cap B) = 0.1323[/tex], find
a.[tex]P(A\cup B)[/tex], and
b. [tex]P(A\cap B)[/tex]


u... what they have in common
n... what neither of them have.


thanks so much!

[Guest]

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

[Guest]

what are you saying