Question

63% of US adults opposed to taxes on junk food and soda. You randomly select 10 US adults. Find the probability that the number of adults who oppose special tax on junk food and soda (a) exactly one, (b) at least four, and (c) less than eight

Answer 1

Answer:

a) 0.0008 = 0.08% probability that the number of adults who oppose special tax on junk food and soda is exactly one.

b) 0.9644 = 96.44% probability that the number of adults who oppose special tax on junk food and soda is at least four.

c) 0.7795 = 77.95% probability that the number of adults who oppose special tax on junk food and soda is less than eight.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they are opposed to taxes on junk food and soda, or they are not. Each adult is independent of other adults, which means that the binomial probability distribution is used to solve the 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.

63% of US adults opposed to taxes on junk food and soda.

This means that $p = 0.63$

You randomly select 10 US adults.

This means that $n = 10$

(a) exactly one

This is $P(X = 1)$. So

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

$P(X = 1) = C_{10,1}.(0.63)^{1}.(0.37)^{9} = 0.0008$

0.0008 = 0.08% probability that the number of adults who oppose special tax on junk food and soda is exactly one.

(b) at least four

This is

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

So

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

$P(X = 0) = C_{10,0}.(0.63)^{0}.(0.37)^{10} \approx 0$

$P(X = 1) = C_{10,1}.(0.63)^{1}.(0.37)^{9} = 0.0008$

$P(X = 2) = C_{10,2}.(0.63)^{2}.(0.37)^{8} = 0.0063$

$P(X = 3) = C_{10,3}.(0.63)^{3}.(0.37)^{7} = 0.0285$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0008 + 0.0063 + 0.0285 = 0.0356$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.0356 = 0.9644$

0.9644 = 96.44% probability that the number of adults who oppose special tax on junk food and soda is at least four.

(c) less than eight

This is

$P(X < 8) = 1 - P(X \geq 8)$

In which

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 8) = C_{10,8}.(0.63)^{8}.(0.37)^{2} = 0.1529$

$P(X = 9) = C_{10,1}.(0.63)^{9}.(0.37)^{1} = 0.0578$

$P(X = 10) = C_{10,2}.(0.63)^{10}.(0.37)^{0} = 0.0098$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1529 + 0.0578 + 0.0098 = 0.2205$

$P(X < 8) = 1 - P(X \geq 8) = 1 - 0.2205 = 0.7795$

0.7795 = 77.95% probability that the number of adults who oppose special tax on junk food and soda is less than eight.