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

James Dale

James Dale

Answered question

2021-12-20

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) P(3)=
(b) P(x4)=
(c) P(x2)=

Answer & Explanation

Ronnie Schechter

Ronnie Schechter

Beginner2021-12-21Added 27 answers

Step 1
Given Data:
The probability that adults say that cashews are their favorite kind of nut is: p=0.43
The total number of adults randomly selected is: n=12.
The probability that the cashew nuts are not their favorite kind of nut is,
q=1p
Substitute values in the above expression.
q=10.43
=0.57
The binomial expression to calculate the probability of x adults who says that cashews are their favorite kind of nut is,
P(x)=nCxpxqnx
Substitute values in the above expression.
P(x)=12Cx×(0.43)x×(0.57)12x
Step 2
(a) The expression to calculate the probability that exactly three adults say that the cashews are their favorite kind of nut is,
P(3)=12C3×(0.43)3×(0.57)123
=220×(0.43)3×(0.57)9
=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.
P(x4)=1(P(0)+P(1)+P(2)+P(3))
Substitute values in the above expression.
P(x4)=1({12}C0×(0.43)0×(0.57)12+12C1×0.43×(0.57)11+12C2×(0.43)2×(0.43)2×(0.57)10+12C3×(0.43)3×(0.57)9)
=1(0.0017+0.0106+0.0441+0.1110)
=10.1674
=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
ambarakaq8

ambarakaq8

Beginner2021-12-22Added 31 answers

Step 1
The following notation varies slightly from book-to-book
binomial formula:
P(x)=(nx)×px×(1p)x
where (nx)=n!x!(nx)!
Here n=12, p=0.42 and 1p=10.42=0.58
a) P(X=3)=12!3!(123)!(0.423)(0.58123)
P(X=3)=220(0.423)(0.589)
P(X=3)=0.1211
b) P(X4)=P(4)+P(5)+docs+P(12)
P(X4)=1P(3)P(2)+P(1)P(0)
P(X=3)=12!3!(123)!(0.423)(0.58123)=0.121066
P(X=2)=12!2!(122)!(0.422)(0.58122)=0.050156
P(X=1)=12!1!(121)!(0.421)(0.58121)=0.012593
P(X=0)=12!0!(120)!(0.420)(0.58120)=0.001449
P(X4)=10.1210060.0501560.0125930.001449
P(X4)=0.8147
c) P(X2)=P(2)+P(1)+P(0)
P(X2)=0.050156+0.012593
nick1337

nick1337

Expert2021-12-28Added 777 answers

Solution:
Given that
p=42%=0.42
q=1P=10.42=0.58
n=12
Using binomial probability formula
p(x=x)=nCxpxqn×x
a) p(x=3)=12C3(0.42)3(0.58)123
=0.121
P(3)=0.121
b) P(x4)=1p(x<4)
=1p(x=0)+p(x=1)+p(x=2)+p(x=3)
=10.0014+0.0126+0.0502+0.1211
=10.1853
P(x4)=0.815
c) P(x2)=p(x=0)+p(x=1)+p(x=2)
=0.0014+0.0126+0.0502
p(x2)=0.064

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