Question

Problem 4: Teen smoking

According to a report published by the Centers for Disease Control and Prevention, about 20 percent of high school students currently use a tobacco product. This number of down from 23 percent in 2014. (https://www.cdc.gov/tobacco/data_statistics/fact_sheets/youth_data/tobacco_use/index.htm). We would like to conduct a study to evaluate high school students’ attitudes toward scenes of smoking in the movies. Suppose you randomly select students to survey them on their opinion about this question.

a) What is the probability that none of the first 4 students you interview is a smoker? (1 point)

b) What is the probability that there are no more than two smokers among 10 students you randomly choose? (2 points)

c) What is the probability that exactly 3 out of a new sample of 10 students do not smoke? (1 point)

Answer 1

We can do these as Binomial probability

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Also sum of probabilities from 0 till n = 1, i.e P(0) + P(1) + P(2) +.......+P(n) = 1

Please note ⁿC_x = n! / [(n-x)!*x!]

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(a) Here n = 4,p = 20% = 0.2, q = 1 - p = 0.8

P(X = 0) = ⁴C₀ * (0.2)⁰ * (0.8)^{4-0 = 4} = 0.4096

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(b) Here n = 10,To find P(No more than 2) = P(X 2) = P(0) + P(1) + P(2)

P(X = 0) = ¹⁰C₀ * (0.2)⁰ * (0.8)^{10-0 = 10} = 0.1074

P(X = 1) = ¹⁰C₁ * (0.2)¹ * (0.8)^{10-1 = 9} = 0.2684

P(X = 2) = ¹⁰C₂ * (0.2)² * (0.8)^{10-2 = 8} = 0.3020

Therefore P(X 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778

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(c) Here n = 10

P(X = 3) = ¹⁰C₃ * (0.2)³ * (0.8)^{10-3 = 7} = 0.2013

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