probability in reverse cocaine sting The American Statistician (May 1991) descri

Question

probability in reverse cocaine sting The American Statistician (May 1991) described an interesting application of a discrete probability distribution in a case involving illegal drugs. It all started with a "bust" in a midsized Florida city. During the bust, police seized 496 foil packets of a white, powdery substance, presumably cocaine. Since it is not a crime to buy or sell non-narcotic cocaine look-a-like (e.g. inert powders), detectives had to prove that the packets contained genuine cocaine in order to convict their suspects of drug trafficking When the police laboratory randomly selected and chemically tested four of the packets, all four tested positive for cocaine. This finding led to the conviction of the traffickers After the conviction, the police decided to use the remaining foil packets (i.e. those not tested) in reverse sting operations. Two of these packets were randomly selected and sold by undercover officers to a buyer Between the sale and the arrest, however, the buyer disposed of the evidence. The key question is, "Beyond reasonable doubt, did the defendant really purchase cocaine?"

Explanation / Answer

Here as sample size is less than 1% of population size, we will take it as binomial distribution.

(a) CHaracteristics of binomial distriution

(i) There must be a fixed number of trials

(ii) Each trial can have only two outcomes

(iii) The outcomes of each trial must be independent of each other

(iv) The probability of a success must remain the same for each trial

(b) Pr(Success) =p^ = 331/496 = 0.667

(c) Pr(failure) = 1 -p^ = 1 - 0.667 = 0.333

(d) Here the probability distribution table

P(X) = 4Cx (p)x(1-p)(4-x) =

(5) All the cocaine powder contains cocaine = Pr(X = 4) = 0.1983

(6) Pr(Two selected randomly donot contain cocaine) = 2C2 * (0.333)2 =  0.111

(7) As events are independent so

Pr(FIrst four Contain Cocaine and Next two doesn't contain cocaine)Pr(FIrst four Contain Cocaine) *Pr(Next two doesn't contain cocaine) = 0.1987 * 0.111 = 0.022

(8) Here the probability of this event happening is not usual and it is less likely to occur, but that doesn't mean we ould not warrant the conviction as overall there are 4 cocaine packets out of 6 packets. So, that's also a strong conviction to occur.

x P(x) 0 0.0122 1 0.0983 2 0.2957 3 0.3955 4 0.1983

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.