To avoid detection at customs, a traveler places 6 narcotic tablets in

kiki195ms

kiki195ms

Answered question

2021-11-14

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 & Explanation

William Yazzie

William Yazzie

Beginner2021-11-15Added 20 answers

In the bottle, there are (6+9)=15 tablets.
Among the 15 tablets, 6 are narcotic tablets and 9 are vitamin tablets.
The traveller will be arrested if any of the selected tablets is a narcotic tablet.
So we have to find the probability that at least one tablet will be narcotic
P(at least one tablet narcotic)
=1-P(no tablets are narcotic)
We can select three nonnarcotic tablets in 9C3 ways
We can select any tablet in 15C3 ways
P(at least one tablet narcotic)
=1-P(no tablets are narcotic)
=19C315C3
=184455
=371455
=5365
Answer: 5365
Nick Camelot

Nick Camelot

Skilled2023-06-17Added 164 answers

Answer:
40273
Explanation:
Given information:
- Total tablets in the bottle: 6 narcotic tablets + 9 vitamin tablets = 15 tablets
- Number of tablets selected for analysis: 3 tablets
We need to calculate the probability of selecting all 3 narcotic tablets out of the 15 tablets.
The probability of selecting the first narcotic tablet is 615 since there are 6 narcotic tablets out of the total 15 tablets.
After selecting the first narcotic tablet, there are 5 narcotic tablets left out of the remaining 14 tablets. So the probability of selecting the second narcotic tablet is 514.
Finally, after selecting the first and second narcotic tablets, there are 4 narcotic tablets left out of the remaining 13 tablets. Therefore, the probability of selecting the third narcotic tablet is 413.
To find the overall probability, we multiply these individual probabilities together since each selection is independent:
P({selecting 3 narcotic tablets})=615×514×413
Simplifying this expression, we get:
P({selecting 3 narcotic tablets})=215×514×413
Calculating this expression, we find:
P({selecting 3 narcotic tablets})=40273
Therefore, the probability that the traveler will be arrested for illegal possession of narcotics is 40273.
madeleinejames20

madeleinejames20

Skilled2023-06-17Added 165 answers

The number of ways to select 3 narcotic tablets from the 6 available is (63).
Therefore, the probability of selecting 3 narcotic tablets out of the 15 tablets is:
P({arrested})=(63)(153)
Simplifying further:
P({arrested})=6!3!3!15!3!12!
Now we can calculate the probability:
P({arrested})=6!·3!12!3!3!·15!
Canceling out the common factors:
P({arrested})=6·5·415·14·13
Calculating further:
P({arrested})=1202730
Thus, the probability that the traveler will be arrested for illegal possession of narcotics is 1202730.
Eliza Beth13

Eliza Beth13

Skilled2023-06-17Added 130 answers

Step 1:
The total number of tablets in the bottle is 6 narcotic tablets + 9 vitamin tablets = 15 tablets.
The probability of selecting a narcotic tablet on the first draw is 615 since there are 6 narcotic tablets out of 15 total tablets.
Step 2:
After the first draw, there will be 14 tablets left in the bottle (5 narcotic tablets and 9 vitamin tablets). The probability of selecting a narcotic tablet on the second draw, given that the first tablet was a narcotic tablet, is 514.
Similarly, after the second draw, there will be 13 tablets left in the bottle (4 narcotic tablets and 9 vitamin tablets). The probability of selecting a narcotic tablet on the third draw, given that the first two tablets were narcotic tablets, is 413.
To find the overall probability of selecting 3 narcotic tablets in a row, we need to multiply the probabilities of each individual draw:
P(NNN)=P(N)×P(N|N)×P(N|N,N)=615×514×413=215
Therefore, the probability that the traveler will be arrested for illegal possession of narcotics is 215.

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