Thursday, July 30, 2009

Probability HELP NOW!!!?

In order to verify the accuracy of their financial accounts, companies use auditors on a regular basis to check the accuracy of the accounts entries. The companies accountants make errors in 5% of the entries. If an auditor randomly checks 3 entries, and the random variable Y is the number of errors detected by the auditor, then





a) Find the probability distribution of the random variable Y,





b) Find E(Y) and Var(Y),





c) And the probability that more than one error will be found by the auditor?





Step by step please.

Probability HELP NOW!!!?
Let X be the number of entries with defects. X has the binomial distribution with n = 3 trials and success probability p = 0.05





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.


Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.





X ~ Binomial( n , p )





the mean of the binomial distribution is n * p = 0.15


the variance of the binomial distribution is n * p * (1 - p) = 0.1425


the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 0.3774917





The Probability Mass Function, PMF,


f(X) = P(X = x) is:





P( X = 0 ) = 0.857375


P( X = 1 ) = 0.135375


P( X = 2 ) = 0.007125


P( X = 3 ) = 0.000125








P( X ≥ 1 ) = 1 - P(X %26lt; 1)


= 1 - P(X = 0)


= 1 - 0.857375


= 0.142625
Reply:When you're asking to solve the Expectation, i.e. E(y), it would be helpful if you could tell which kind of distribution this is.





Each distribution has different set of rules for E(y) and Var(Y).





If you don't know what kind of distribution this is, there is not much anyone can do.
Reply:Did you check the answer to part C for the last time you asked this question? This is a binomial random variable





The probability distribution for Y = x, 0 %26lt;= x %26lt;=3, is





P{Y = x} = C(3, x) (1/20)^x (19/20)^(3 - x).





So,





P(0) = (1) (1) (19/20)^3 = 6859/8000


P(1) = (3) (1/20) (19/20)^2 = 1083/8000


P(2) = (3) (1/20)^2 (19/20) = 57/8000


P(3) = (1) (1/20)^3 (1) = 1/8000





As you can see, Σ P(n) = 1.





(b) Y is a binomial random variable with parameters n = 3 and p = 1/20.





E[Y] = np = 3/20.


Var(Y) = np(1 - p) = (3/20)(19/20) = 57/400





(c) Refer to the link provided for a step-by-step solution to:





P(2) + P(3) = 29/400





--





OK, LOL. Sometimes I'm not sure that people are actually reading their answers, and I really do try to give good ones.





So, OK. Since it's a binomial random variable with parameters n = 3 and p = 5% = 1/20, (a) and (b) practically write themselves (see above.)





Basically, since this is a discrete random variable taking on only 4 values {0, 1, 2, 3}, you can simply compute each probability separately, once you understand the underlying concept, which is this:





If the probability of finding an error is 1/20, the probability of finding no error is 1 - 1/20 = 19/20.





And, the accuracy of any particular entry is independent of the accuracy of any other entry. So the probability of, say, getting an error on the first entry and no error on the second entry is:





(1/20)(19/20) = 19/400.





If, out of 3 tries, an auditor finds 1 error, he may find it in any of 3 choose 1 = 3 ways.





So the probability of finding 1 error is





(3) (1/20) (19/20) (19/20) = (3) (1/20)^1 (19/20)^2, which is just another way of saying





(1/20) (19/20) (19/20)


+ (19/20) (1/20) (19/20)


+ (19/20) (19/20) (1/20)





The same goes for any other probability Y = n; it's just





(3 choose n) times the probability of finding n errors times the probability of finding (n - 3) correct entries, or





(3 choose n) (1/20)^n (19/20)^(3 - n)


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