To avoid detection at customs, a traveler places

Lester John Sus

Lester John Sus

Answered question

2022-05-03

To avoid detection at customs, a traveler places 6 narcotic tablets in a bottle 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 the traveler will be arrested for illegal possession of narcotics?

Answer & Explanation

karton

karton

Expert2022-07-07Added 613 answers

A traveler places narcotic tablets in a bottle with 9 vitamin tablets that are similar in appearance (so, there will be total 6+9=15 tablets in the bottle).
The customs official selects 3 of the tablets at random for analysis.
* We need to find the probability that the traveler will be arrested for illegal possession of narcotics.
The traveler will be arrested if the customs official finds at least 1 narcotic table among 3 randomly selected tablets from the traveler's bottle.
Let random variable X represent the number of narcotic tablets among 3 randomly selected tablets from the traveler's bottle. Therefore, we need to find P(X1).
Lets consider as a success an event when the customs official finds the narcotic table.
So, we have a random sample of size n=3 (the custom official selects 3 tablets for
analysis), taken from N=6+9=15 items (there are 15 tables in the traveler's bottle) of which k=6 are labeled success (there are 6 narcotic tablets in the bottle) and N-k=15-6=9 are labeled failure.

Thus, X is hypergeometric random variable.

The probability distribution of hypergeometric random variable X is:
h(x;N,n,k)=(kx)(N-kn-x)(Nn)


We need to find

P (X  1) = 1  P (X = 0) = 1  h(0; 15, 3, 6)                 (1)
According to probability distribution of the hypergeometric random variable X, we
have:
h(0; 15, 3, 6) =(60)(15-63-0)(153)
=(60)(93)(153)=6!0!(6-0)!9!3!(9-3)!15!3!(15-3)!
=(1)(84)455 0.1846              (2)

According equations (1) and (2), we get the required probability as follows:
P (X  1) = 1  h(0; 15, 3, 6) = 1  0.1846 = 0.8154
So, the probability that the traveler will be arrested is 0.8154

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