Question

36​% of adults say cashews are their favorite kind of nut. you randomly select 12 adults and ask each to name his or her favorite nut. find the probability that the number who say cashews are their favorite nut is​ (a) exactly​ three, (b) at least​ four, and​ (c) at most two. if​ convenient, use technology to find the probabilities

Answer 1

Answer:

a) 18.49% probability that the number who say cashews are their favorite nut is​ exactly three.

b) 67.99% probability that the number who say cashews are their favorite nut is​ at least four.

c) 13.52% probability that the number who say cashews are their favorite nut is at most two.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either cashews are their favorite kind of nut, or they are not. The probability of an adult having cashews as their favorite kind of nut is independent of other adults. So we use the binomial probability distribution 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.

36​% of adults say cashews are their favorite kind of nut.

This means that $p = 0.36$

12 adults

This means that $n = 12$

(a) exactly​ three

This is P(X = 3).

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

$P(X = 3) = C_{12,3}.(0.36)^{3}.(0.64)^{9} = 0.1849$

18.49% probability that the number who say cashews are their favorite nut is​ exactly three.

(b) at least​ four

Either there are less than four, or there are at least four. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \geq 4)$. So

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

In which

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

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

$P(X = 0) = C_{12,0}.(0.36)^{0}.(0.64)^{12} = 0.0047$

$P(X = 1) = C_{12,1}.(0.36)^{1}.(0.64)^{11} = 0.0319$

$P(X = 2) = C_{12,2}.(0.36)^{2}.(0.64)^{10} = 0.0986$

$P(X = 3) = C_{12,3}.(0.36)^{3}.(0.64)^{9} = 0.1849$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0047 + 0.0319 + 0.0986 + 0.1849 = 0.3201$

$P(X \geq 4) = 1 - P(X < 4)[ = 1 - 0.3201 = 0.6799$

67.99% probability that the number who say cashews are their favorite nut is​ at least four.

c) At most two

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{12,0}.(0.36)^{0}.(0.64)^{12} = 0.0047$

$P(X = 1) = C_{12,1}.(0.36)^{1}.(0.64)^{11} = 0.0319$

$P(X = 2) = C_{12,2}.(0.36)^{2}.(0.64)^{10} = 0.0986$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0047 + 0.0319 + 0.0986 = 0.1352$

13.52% probability that the number who say cashews are their favorite nut is at most two.

Answer 2

Its 36% of 12
36/100×12= 4.32
so it atleast four, which is b.
Think that may help