Question

To avoid detection at customs, a traveler has placed 6 narcotic tablets in an Urn containing 9 vitamin pills that are similar in appearance. If the customs official selects 3 of the tablets at random for analysis, what is the probability that that the traveler will be arrested for illegal possession of narcotics?

Answer 1

`(4)/(91)`

Step-by-step explanation:

We are told that to avoid detection at customs, a traveler has placed 6 narcotic tablets in an Urn containing 9 vitamin pills that are similar in appearance.

To find the probability that the traveler will be arrested for illegal possession of narcotics we will use theoretical probability formula.

`\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}`

Favorable outcomes will be
`_(3)^(6)\textrm{C}`.

`\text{Favorable outcomes}=_(3)^(6)\textrm{C}=(6!)/((6-3)!3!)`

`\text{Favorable outcomes}=(6!)/(3!3!)`

`\text{Favorable outcomes}=(6\cdot 5\cdot 4\cdot3!)/(3\cdot 2\cdot 1\cdot3!)`

`\text{Favorable outcomes}=(6\cdot 5\cdot 4)/(3\cdot 2)`

`\text{Favorable outcomes}=5\cdot 4=20`

Now let us find number of possible outcomes.

`\text{Number of possible outcomes}=_(3)^(15)\textrm{C}=(15!)/((15-3)!3!)`

`\text{Number of possible outcomes}=(15!)/(12!3!)`

`\text{Number of possible outcomes}=(15\cdot 14\cdot 13\cdot 12!)/(12!\cdot 3\cdot 2\cdot 1)`

`\text{Number of possible outcomes}=(15\cdot 14\cdot 13)/( 3\cdot 2\cdot 1)`

`\text{Number of possible outcomes}=5\cdot 7\cdot 13=455`

Upon substituting our values in probability formula we will get,

`\text{Probability that the traveler will be arrested for illegal possession of narcotics}=(20)/(455)`

Upon dividing numerator and denominator by 5 we will get,

`\text{Probability that the traveler will be arrested for illegal possession of narcotics}=(4)/(91)`

Therefore, the probability that the traveler will be arrested for illegal possession of narcotics is
`(4)/(91)`.