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

James Dale 2021-12-20 Answered
42​% 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.
(a) \(\displaystyle{P}{\left({3}\right)}=\)
(b) \(\displaystyle{P}{\left({x}\geq{4}\right)}=\)
(c) \(\displaystyle{P}{\left({x}\le{2}\right)}=\)

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Expert Answer

Ronnie Schechter
Answered 2021-12-21 Author has 54 answers
Step 1
Given Data:
The probability that adults say that cashews are their favorite kind of nut is: \(\displaystyle{p}={0.43}\)
The total number of adults randomly selected is: \(\displaystyle{n}={12}\).
The probability that the cashew nuts are not their favorite kind of nut is,
\(\displaystyle{q}={1}-{p}\)
Substitute values in the above expression.
\(\displaystyle{q}={1}-{0.43}\)
\(\displaystyle={0.57}\)
The binomial expression to calculate the probability of 'x' adults who says that cashews are their favorite kind of nut is,
\(\displaystyle{P}{\left({x}\right)}=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{q}^{{{n}-{x}}}\)
Substitute values in the above expression.
\(\displaystyle{P}{\left({x}\right)}=^{{{12}}}{C}_{{{x}}}\times{\left({0.43}\right)}^{{{x}}}\times{\left({0.57}\right)}^{{{12}-{x}}}\)
Step 2
(a) The expression to calculate the probability that exactly three adults say that the cashews are their favorite kind of nut is,
\(\displaystyle{P}{\left({3}\right)}=^{{{12}}}{C}_{{{3}}}\times{\left({0.43}\right)}^{{{3}}}\times{\left({0.57}\right)}^{{{12}-{3}}}\)
\(\displaystyle={220}\times{\left({0.43}\right)}^{{{3}}}\times{\left({0.57}\right)}^{{{9}}}\)
\(\displaystyle={0.1111}\)
Thus, the probability that exactly three adults say that the cashews are their favorite kind of nut is 0.111.
(b) The expression to calculate the probability that at least four of the adults like cashews nuts.
\(\displaystyle{P}{\left({x}\geq{4}\right)}={1}-{\left({P}{\left({0}\right)}+{P}{\left({1}\right)}+{P}{\left({2}\right)}+{P}{\left({3}\right)}\right)}\)
Substitute values in the above expression.
\(\displaystyle{P}{\left({x}\geq{4}\right)}={1}-{\left(^{\left\lbrace{12}\right\rbrace}{C}_{{{0}}}\times{\left({0.43}\right)}^{{{0}}}\times{\left({0.57}\right)}^{{{12}}}+^{{{12}}}{C}_{{{1}}}\times{0.43}\times{\left({0.57}\right)}^{{{11}}}+^{{{12}}}{C}_{{{2}}}\times{\left({0.43}\right)}^{{{2}}}\times{\left({0.43}\right)}^{{{2}}}\times{\left({0.57}\right)}^{{{10}}}+^{{{12}}}{C}_{{{3}}}\times{\left({0.43}\right)}^{{{3}}}\times{\left({0.57}\right)}^{{{9}}}\right)}\)
\(\displaystyle={1}-{\left({0.0017}+{0.0106}+{0.0441}+{0.1110}\right)}\)
\(\displaystyle={1}-{0.1674}\)
\(\displaystyle={0.8326}\)
Thus, the probability that at least four adults like cashews is0.8326.
Step 3
(c) The expression to calculate the probability that at most two adults' like cashews is,
\(\displaystyle{P}{\left({x}\le{2}\right)}={P}{\left({0}\right)}+{P}{\left({1}\right)}+{P}{\left({2}\right)}\)
Substitute values in the above expression.
\(\displaystyle{P}{\left({x}\le{2}\right)}=^{{{12}}}{C}_{{{0}}}\times{\left({0.43}\right)}^{{{0}}}\times{\left({0.57}\right)}^{{{12}}}+^{{{12}}}{C}_{{{1}}}\times{\left({0.43}\right)}\times{\left({0.57}\right)}^{{{11}}}+^{{{12}}}{C}_{{{2}}}\times{\left({0.43}\right)}^{{{2}}}\times{\left({0.57}\right)}^{{{10}}}\)
\(\displaystyle={0.0017}+{0.0106}+{0.0441}\)
\(\displaystyle={0.0564}\)
Thus, the probability that at most two adults like cashews is 0.0564.
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ambarakaq8
Answered 2021-12-22 Author has 1496 answers
Step 1
The following notation varies slightly from book-to-book
binomial formula:
\(\displaystyle{P}{\left({x}\right)}={\left(\begin{array}{c} {n}\\{x}\end{array}\right)}\times{p}^{{{x}}}\times{\left({1}-{p}\right)}^{{{x}}}\)
where \(\displaystyle{\left(\begin{array}{c} {n}\\{x}\end{array}\right)}={\frac{{{n}!}}{{{x}!{\left({n}-{x}\right)}!}}}\)
Here \(\displaystyle{n}={12},\ {p}={0.42}\) and \(\displaystyle{1}-{p}={1}-{0.42}={0.58}\)
a) \(\displaystyle{P}{\left({X}={3}\right)}={\frac{{{12}!}}{{{3}!{\left({12}-{3}\right)}!}}}{\left({0.42}^{{{3}}}\right)}{\left({0.58}^{{{12}-{3}}}\right)}\)
\(\displaystyle{P}{\left({X}={3}\right)}={220}{\left({0.42}^{{{3}}}\right)}{\left({0.58}^{{{9}}}\right)}\)
\(\displaystyle{P}{\left({X}={3}\right)}={0.1211}\)
b) \(\displaystyle{P}{\left({X}\geq{4}\right)}={P}{\left({4}\right)}+{P}{\left({5}\right)}+{d}{o}{c}{s}+{P}{\left({12}\right)}\)
\(\displaystyle{P}{\left({X}\geq{4}\right)}={1}-{P}{\left({3}\right)}-{P}{\left({2}\right)}+{P}{\left({1}\right)}-{P}{\left({0}\right)}\)
\(\displaystyle{P}{\left({X}={3}\right)}={\frac{{{12}!}}{{{3}!{\left({12}-{3}\right)}!}}}{\left({0.42}^{{{3}}}\right)}{\left({0.58}^{{{12}-{3}}}\right)}={0.121066}\)
\(\displaystyle{P}{\left({X}={2}\right)}={\frac{{{12}!}}{{{2}!{\left({12}-{2}\right)}!}}}{\left({0.42}^{{{2}}}\right)}{\left({0.58}^{{{12}-{2}}}\right)}={0.050156}\)
\(\displaystyle{P}{\left({X}={1}\right)}={\frac{{{12}!}}{{{1}!{\left({12}-{1}\right)}!}}}{\left({0.42}^{{{1}}}\right)}{\left({0.58}^{{{12}-{1}}}\right)}={0.012593}\)
\(\displaystyle{P}{\left({X}={0}\right)}={\frac{{{12}!}}{{{0}!{\left({12}-{0}\right)}!}}}{\left({0.42}^{{{0}}}\right)}{\left({0.58}^{{{12}-{0}}}\right)}={0.001449}\)
\(\displaystyle{P}{\left({X}\geq{4}\right)}={1}-{0.121006}-{0.050156}-{0.012593}-{0.001449}\)
\(\displaystyle{P}{\left({X}\geq{4}\right)}={0.8147}\)
c) \(\displaystyle{P}{\left({X}\le{2}\right)}={P}{\left({2}\right)}+{P}{\left({1}\right)}+{P}{\left({0}\right)}\)
\(\displaystyle{P}{\left({X}\le{2}\right)}={0.050156}+{0.012593}+{0.001449}\)
\(\displaystyle{P}{\left({X}\le{2}\right)}={0.0642}\)
0
nick1337
Answered 2021-12-28 Author has 10701 answers

Solution:
Given that
\(p=42\%=0.42\)
\(q=1-P=1-0.42=0.58\)
\(n=12\)
Using binomial probability formula
\(p(x=x)=^{n}C_{x}p^{x}q^{n\times x}\)
a) \(p(x=3)=^{12}C_{3}(0.42)^{3}(0.58)^{12-3}\)
\(=0.121\)
\(P(3)=0.121\)
b) \(P(x\geq4)=1-p(x<4)\)
\(=1-p(x=0)+p(x=1)+p(x=2)+p(x=3)\)
\(=1-0.0014+0.0126+0.0502+0.1211\)
\(=1-0.1853\)
\(P(x\geq4)=0.815\)
c) \(P(x\le2)=p(x=0)+p(x=1)+p(x=2)\)
\(=0.0014+0.0126+0.0502\)
\(p(x\le2)=0.064\)

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