According to USA Today (March 18, 1997), of 4 million workers in the general workforce, 5.8% tested positive for drugs. Of those testing positive, 22.5% were cocaine users and 54.4% marijuana users. (a) What is the probability that of 10 workers testing positive, 2 are cocaine users, 5 are marijuana users, and 3 are users of other drugs? (b) What is the probability that of 10 workers testing positive, all are marijuana users? (c) What is the probability that of 10 workers testing positive, none is a cocaine user?

To solve this question, we shall apply the Multinomial and the Binomial distribution.

We are given that,

5.8% of the workers are positive

Of those testing positive,

22.5% were cocaine users

54.4% marijuana users.

Of those who are positive, the percentage using other drugs is, 100%-(22.5%+54.4%)=23.1%

Let the event $C$ represent workers who are cocaine users.

Let the event $M$ represent workers who are marijuana users.

Let the event $O$ represent workers who use other drugs.

Now, $p(C)=0.225$,$p(M)=0.544$, $p(O)=0.231$

$a)$

Of 10 workers testing positive, the probability that 2 are cocaine users, 5 are marijuana users, and 3 are users of other drugs is given as,

$\binom{10}{2!5!3!}0.225^20.544^50.231^3=2520\times 0.050625\times 0.047642495\times0.012326391=0.075$

$b)$

The probability that of 10 workers testing positive, all are marijuana users.

$\binom{10}{0!10!0!}0.225^00.544^{10}0.231^0=0.00227$

$c)$

The probability that of 10 workers testing positive, none is a cocaine user is obtained using the Binomial distribution as follows.

$n=10,p=0.225, x=0$

Therefore,

$p(x=0)=\binom{10}{0}0.225^00.775^{10}=0.0782.$