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lesya692 [45]
3 years ago
14

Twenty identical looking packets of white power are such that 15 contain cocaine and 5 do not. Four packets were randomly select

ed, and the contents were tested and found to contain cocaine. Two additional packets were selected from the remainder and sold by undercover police officers to a single buyer. What is the probability that the 6 packets randomly selected are such that the first 4 all contain cocaine and the 2 sold to the buyer do not?
Mathematics
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

The probability of selecting randomly 6 packets such that the first 4 contain cocaine and 2 not is 0.3522.

Step-by-step explanation:

Probabilities of selecting 4 packets with the ilegal substance:

(\frac{15}{4})\\= \frac{15!}{4!(15-4)!}

Combinassions possible= 1365

Probabilities of selecting 2 packets with white powder:

(\frac{5}{2})= \frac{5!}{2!(5-2)!}

Combinations possible= 10

Probabilities of selecting 6 packets from the totality of them:

(\frac{20}{6}) = \frac{20!}{6!(20-6)!}

Combinations possible= 38760

The probability of picking 4 with the substance and 2 with only white powder is:

\frac{(15ncr4)(5ncr2)}{(20ncr6)} = 0.3522

I hope this answer helps you.

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There would be 173,535 lionfish after 6 years.

Since lionfish are considered an invasive species, with an annual growth rate of 67%, ya scientist estimates there are 8,000 lionfish in a certain bay after the first year, A) to write the explicit equation for f (n) that represents the number of lionfish in the bay after n years; B) determine how many lionfish will be in the bay after 6 years; and C) if scientists remove 1,200 fish per year from the bay after the first year, determine what is the recursive equation for f (n); the following calculations must be performed:

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Therefore, there would be 173,535 lionfish after 6 years.

Learn more about maths in brainly.com/question/25851847

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