Question

3.10 Shuster (1991) describes a number of probability calculations that he did for a court case involving the sale of cocaine. A Florida police department seized 496 suspected packets of cocaine, of which four were randomly selected and tested and found to actually be cocaine. The police then chose two more packets at random and, posing as drug dealers, sold the packets to the defendant. These last two packets were lost before they could be tested to verify that they were, indeed, cocaine. (a) If the original 496 packets were composed of N packets of cocaine and M = 496-N noncocaine, show that the probability of selecting 4 cocaine packets and then 2 noncocaine packets, which is the probability that the defendant is innocent of buying cocaine, is $inom{N}{4} inom{M}{2} over inom{N+M}{4} inom{N+M-4}{2}$ (b) Maximizing (in M and N) the probability in part (a) maximizes the defendant's ""innocence probability"". Show that this probability is .022, attained at M = 165 and N = 331.

          3.10 Shuster (1991) describes a number of probability calculations that he did for a court
case involving the sale of cocaine. A Florida police department seized 496 suspected
packets of cocaine, of which four were randomly selected and tested and found to
actually be cocaine. The police then chose two more packets at random and, posing
as drug dealers, sold the packets to the defendant. These last two packets were lost
before they could be tested to verify that they were, indeed, cocaine.
(a) If the original 496 packets were composed of N packets of cocaine and M = 496-N
noncocaine, show that the probability of selecting 4 cocaine packets and then 2
noncocaine packets, which is the probability that the defendant is innocent of
buying cocaine, is
$inom{N}{4} inom{M}{2} over inom{N+M}{4} inom{N+M-4}{2}$
(b) Maximizing (in M and N) the probability in part (a) maximizes the defendant's
""innocence probability"". Show that this probability is .022, attained at M = 165
and N = 331.

3.10 Shuster (1991) describes a number of probability calculations that he did for a court
case involving the sale of cocaine. A Florida police department seized 496 suspected
packets of cocaine, of which four were randomly selected and tested and found to
actually be cocaine. The police then chose two more packets at random and, posing
as drug dealers, sold the packets to the defendant. These last two packets were lost
before they could be tested to verify that they were, indeed, cocaine.
(a) If the original 496 packets were composed of N packets of cocaine and M = 496-N
noncocaine, show that the probability of selecting 4 cocaine packets and then 2
noncocaine packets, which is the probability that the defendant is innocent of
buying cocaine, is
inomN4 inomM2 over inomN+M4 inomN+M-42
(b) Maximizing (in M and N) the probability in part (a) maximizes the defendant's
""innocence probability"". Show that this probability is .022, attained at M = 165
and N = 331.

Added by Lucas H.

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