Mathematics, 28.05.2021 02:00 Geo777

According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded. a) The probability that at least 2 of them started smoking before 21 years of age is b) The probability that at most 4 of them started smoking before 21 years of age is c) The probability that exactly 3 of them started smoking before 21 years of age is

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Answer from: keke6361

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%.

Answer from: Quest

they disappear if your absent from to long. i found out that the hard way.

step-by-step explanation:

brainliest lol i just need one more brainliest.

Answer from: Quest

step-by-step explanation:

*calories burned = (weight)(calorie factor)*

calories burned = 180 (0.65)

calories burned = 117

the calories he used up is 117, which is 65% of his weight.

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According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years...