Respuesta :
Answer:
a [tex]\mathbf{P(X=3)=0.1406}[/tex]
b [tex]\mathbf{\[P\left( {X = 3 \ or \ 4 } \right) = 0.1025}[/tex]
c [tex]\mathbf{\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = 0.1035}[/tex]
Step-by-step explanation:
Given that:
In a multiple choice quiz:
there are 5 questions
and 4 choices for each question (a, b, c, d)
let X be the correctly answered question = 1 answer only
and Y be the choices for each question = 4 choices
The probability that Robin guessed the correct answer is:
Probability = n(X)/n(Y)
Probability = 1//4
Probability = 0.25
The probability mass function is :
[tex]P(X=x)=0.25 (1-0.25)^{x-1}[/tex]
We are to find the required probability that the first question she gets right is the 3rd question.
i.e when x = 3
[tex]P(X=3)=0.25 (1-0.25)^{3-1}[/tex]
[tex]P(X=3)=0.25 (0.75)^{2}[/tex]
[tex]\mathbf{P(X=3)=0.1406}[/tex]
b) Find the probability that She gets exactly 3 or exactly 4 questions right
we know that :
n = 5 questions
Probability P =0.25
Let represent X to be the number of questions guessed correctly i,e 3 or 4
Then; the probability mass function can be written as:
[tex]\[P\left( {X = x} \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\x\\\end{array}} \right){\left( {0.25} \right)^x}{\left( {1 - 0.25} \right)^{5 - x}}\][/tex]
[tex]P(X = 3 \ or \ 4)= P(X =3) +P(X =4)[/tex]
[tex]\[P\left( {X = 3 \ or \ 4 } \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\3\\\end{array}} \right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^{5 - 3}}\] + \left( {\begin{array}{*{20}{c}}\\5\\\\4\\\end{array}} \right){\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^{5 - 4}}\][/tex]
[tex]\[P\left( {X = 3 \ or \ 4 } \right) = \dfrac{5!}{3!(5-3)!}\right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^{5 - 3}}\] + \dfrac{5!}{4!(5-4)!} \right){\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^{5 - 4}}\][/tex]
[tex]\[P\left( {X = 3 \ or \ 4 } \right) = \dfrac{5!}{3!(2)!}\right){\left( {0.25} \right)^3}{\left( {0.75} \right)^{2}}\] + \dfrac{5!}{4!(1)!} \right){\left( {0.25} \right)^4}{\left( {0.75} \right)^{1}}\][/tex]
[tex]\[P\left( {X = 3 \ or \ 4 } \right) = 0.0879+0.0146[/tex]
[tex]\mathbf{\[P\left( {X = 3 \ or \ 4 } \right) = 0.1025}[/tex]
c) Find the probability if She gets the majority of the questions right.
We know that the probability mass function is :
[tex]\[P\left( {X = x} \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\x\\\end{array}} \right){\left( {0.25} \right)^x}{\left( {1 - 0.25} \right)^{5 - x}}\][/tex]
So; of She gets majority of her answers right ; we have:
The required probability is,
[tex]P(X>2) = P(X=3) +P(X=4) + P(X=5)[/tex]
∴
[tex]\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\3\\\end{array}} \right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^{5 - 3}}\] + \left( {\begin{array}{*{20}{c}}\\5\\\\4\\\end{array}} \right){\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^{5 - 4}}\]+ \left( {\begin{array}{*{20}{c}}\\5\\\\5\\\end{array}} \right){\left( {0.25} \right)^5}{\left( {1 - 0.25} \right)^{5 - 5}}\][/tex]
[tex]\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = \dfrac{5!}{3!(5-3)!}\right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^{5 - 3}}\] + \dfrac{5!}{4!(5-4)!} \right){\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^{5 - 4}}\] + \dfrac{5!}{5!(5-5)!} \right){\left( {0.25} \right)^5}{\left( {1 - 0.25} \right)^{5 - 5}}\][/tex]
[tex]\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = \dfrac{5!}{3!(5-3)!}\right){\left( {0.25} \right)^3}{\left( {1 - 0.75} \right)^{2}}\] + \dfrac{5!}{4!(5-4)!} \right){\left( {0.25} \right)^4}{\left( {0.75} \right)^{1}}\] + \dfrac{5!}{5!(5-5)!} \right){\left( {0.25} \right)^5}{\left( {0.75} \right)^{0}}\][/tex]
[tex]\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = 0.0879 + 0.0146 + 0.001[/tex]
[tex]\mathbf{\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = 0.1035}[/tex]
