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Nesterboy [21]
4 years ago
14

Can you help me with that

Mathematics
1 answer:
pogonyaev4 years ago
8 0
I hope this helps you

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According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se
belka [17]

Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

50% of adult smokers started smoking before 21 years old.

This means that p = 0.5

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that n = 5.

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125

P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625

P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

P(X \leq 4) = 1 - P(X = 5)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125

P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

4 0
3 years ago
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