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3 years ago

8

A box filled with 6 blue cards 6 brown carts and 6 green cards a card is chosen at random from the box what is the probability t

hat it is a blue or green card write your answer as a fraction in simplest form

1 answer:

horsena [70]3 years ago

5 0

2/3 because there are 18 cards together and 12 of them are blue and green so the fraction is 12/18

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2 years ago

Raoul estimated the difference of 97.23 and 8.3 by rounding each number to the nearest whole. What was Raoul's estimate, and wha

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3 years ago

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

2 years ago