Question

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

Answer 1

The probability is 0.3, or 30%.

These are not independent events; one pill being chosen will affect the probability after that, as the pill will not be replaced before selecting the next one.

The probability of getting exactly 1 narcotic pill is given by:
(6/15)(9/14)(8/13) = 432/2730.  It does not matter what order the narcotic pill is in, the overall product will be the same.

The probability of getting exactly 2 narcotic pills is given by:
(6/15)(5/14)(9/13) = 270/2730.  Again, the order these are found in does not matter, as it is multiplication and will not change the product.

The probability of all 3 pills being narcotics is given by:
(6/15)(5/14)(4/13) = 120/2730.

Adding these three possibilities together, we have 822/2730 = 0.30.

Answer 2

The probability that the traveler will be arrested for illegal possession of narcotics $\boxed{0.738}.$

Further Explanation:

The formula of permutation can be expressed as,

$\boxed{^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}}$

Here, “n” represents the total observations and “r” represents number of the observation has to be arranged.

Given:

There are 6 narcotic tablets in a bottle.

There are 9 vitamin tablets in a bottle.

Explanation:

The total number of tablets are $15{\text{ tablets}}.$

The probability of getting a narcotic tablet is $\dfrac{2}{5}.$

The probability of getting a vitamin tablet is $\dfrac{3}{5}.$

The probability that the traveler will be arrested for illegal possession of narcotics can be obtained as follows,

$\begin{aligned}{\text{Probability}} &= {}^3{C_1}{\left( {\frac{2}{5}} \right)^1}{\left( {\frac{3}{5}} \right)^2} + {}^3{C_2}{\left( {\frac{2}{5}} \right)^2}\left( {\frac{3}{5}} \right) + {}^3{C_3}{\left( {\frac{2}{5}} \right)^3}\\&= 0.432 + 0.288 + 0.018\\&= 0.738\\\end{aligned}$

The probability that the traveler will be arrested for illegal possession of narcotics $\boxed{0.738}.$

Learn more:

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2. Learn more about standard normal distribution brainly.com/question/13006989
3. Learn more about confidence interval of meanhttps://brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter: Binomial Distribution

Keywords: detection, customs, traveler, places, 6 narcotic, narcotic tablets, bottle, vitamin, 9 vitamin tablets, similar, customs, official, selects, illegal possession, arrested, traveler, random analysis, probability, narcotics.