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

$N=$Total number of tablets placed in the bottle$=9+6=15$

$m=$ Total number of narcotic drugs in the bottle$=6$

$n=$ Total number of tablets selected randomly by the customs official$=3$

Let $x=$The number of narcotic tablet(s) selected in the sample.

Since the selections are made without replacement hence, the distribution of $x$ would be:

$x$ $Hypergeometric (N=15, m=6, n=3)$

The proof of $x$ becomes:

$P(x=x)=\frac{\begin{pmatrix}m\\ x\end{pmatrix}\begin{pmatrix}N-m\\ \:n-x\end{pmatrix}}{\begin{pmatrix}N\\ \:n\end{pmatrix}}\:\:x=max\left(0,\:n+m-N\right),\:\:min\left(n,\:m\right)$

$Hence,\:max\left(0,\:n+m-N\right)\:and\:min\left(n,m\right)$

$=max\left(0,\:3+6-15\right)\:and\:min\left(3,\:6\right)$

$=max\left(0,\:-6\right)=0\:and\:min=3$

The proof of $x$ becomes:

$\:P\left(x=x\right)=\frac{\begin{pmatrix}6\\ x\end{pmatrix}\begin{pmatrix}9\\ 3-x\end{pmatrix}}{\begin{pmatrix}15\\ 3\end{pmatrix}};\:x=0,\:1,\:2,\:3$

Therefore we need,

$P\left(The\:travellers\:will\:be\:arrested\:for\:illegal\:possesion\:of\:narcortics\right)=$

$P\left(There\:will\:be\:at\,least\:one\:narcotic\:tablet\:in\:the\:sample\right)=$

$P\left(x\ge 1\right)=$

$1-P\left(x<1\right)$ $\left(Law\:of\:complement\:events\right)=$

$1-P\left(x=0\right)$

$=1-\frac{\begin{pmatrix}6\\ 0\end{pmatrix}\begin{pmatrix}9\\ 3\end{pmatrix}}{\begin{pmatrix}15\\ 3\end{pmatrix}}\:\:\left(By\:the\:proof\:of\:x\right)$ $\approx$ $0.8154$