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

Answer:

$\frac{4}{91}$

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 $_{3}^{6}\textrm{C}$.

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

$\text{Favorable outcomes}=\frac{6!}{3!3!}$

$\text{Favorable outcomes}=\frac{6\cdot 5\cdot 4\cdot3!}{3\cdot 2\cdot 1\cdot3!}$

$\text{Favorable outcomes}=\frac{6\cdot 5\cdot 4}{3\cdot 2}$

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

Now let us find number of possible outcomes.

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

$\text{Number of possible outcomes}=\frac{15!}{12!3!}$

$\text{Number of possible outcomes}=\frac{15\cdot 14\cdot 13\cdot 12!}{12!\cdot 3\cdot 2\cdot 1}$

$\text{Number of possible outcomes}=\frac{15\cdot 14\cdot 13}{ 3\cdot 2\cdot 1}$

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

Upon substituting our values in probability formula we will get,

$\text{Probability that the traveler will be arrested for illegal possession of narcotics}=\frac{20}{455}$

Upon dividing numerator and denominator by 5 we will get,

$\text{Probability that the traveler will be arrested for illegal possession of narcotics}=\frac{4}{91}$

Therefore, the probability that the traveler will be arrested for illegal possession of narcotics is $\frac{4}{91}$.