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

Answer: $\dfrac{26}{27}$

Step-by-step explanation:

Given : To avoid detection at customs, a traveler places 6 narcotic tablets in a bottle containing 9 vitamin tablets that are similar in appearance.

Proportion of success : $p=\dfrac{6}{9}=\dfrac{2}{3}$

Sample size taken by customs official : n= 3

Let x be a binomial variable that represents the tablets in the bottle.

Using Binomial probability formula :-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

The probability that the traveler will be arrested for illegal possession of narcotics =$P(x\geq1)=1-P(x=0)$

$1-^3C_0(\dfrac{2}{3})^0(1-\dfrac{2}{3})^3\\\\=1-(1)(1)(\dfrac{1}{3})^3\ [\becuase ^nC_0=1]\\\\=1-\dfrac{1}{27}=\dfrac{26}{27}$

Hence, the probability that the traveler will be arrested for illegal possession of narcotics = $\dfrac{26}{27}$