Question

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. Find the probabilities of each of the following events:

a. The first question she gets right is the 3rd question?
b. She gets exactly 3 or exactly 4 questions right?
c. She gets the majority of the questions right?

Answer 1

a
`\mathbf{P(X=3)=0.1406}`

b
`\mathbf{\[P\left( {X = 3 \ or \ 4 } \right) = 0.1025}`

c
`\mathbf{\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = 0.1035}`

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 :

`P(X=x)=0.25 (1-0.25)^(x-1)`

We are to find the required probability that the first question she gets right is the 3rd question.

i.e when x = 3

`P(X=3)=0.25 (1-0.25)^(3-1)`

`P(X=3)=0.25 (0.75)^(2)`

`\mathbf{P(X=3)=0.1406}`

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:

`\[P\left( {X = x} \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\x\\\end{array}} \right){\left( {0.25} \right)^x}{\left( {1 - 0.25} \right)^(5 - x)}\]`

`P(X = 3 \ or \ 4)= P(X =3) +P(X =4)`

`\[P\left( {X = 3 \ or \ 4 } \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\3\\\end{array}} \right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^(5 - 3)}\] + \left( {\begin{array}{*{20}{c}}\\5\\\\4\\\end{array}} \right){\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^(5 - 4)}\]`

`\[P\left( {X = 3 \ or \ 4 } \right) = (5!)/(3!(5-3)!)\right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^(5 - 3)}\] + (5!)/(4!(5-4)!) \right){\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^(5 - 4)}\]`

`\[P\left( {X = 3 \ or \ 4 } \right) = (5!)/(3!(2)!)\right){\left( {0.25} \right)^3}{\left( {0.75} \right)^(2)}\] + (5!)/(4!(1)!) \right){\left( {0.25} \right)^4}{\left( {0.75} \right)^(1)}\]`

`\[P\left( {X = 3 \ or \ 4 } \right) = 0.0879+0.0146`

`\mathbf{\[P\left( {X = 3 \ or \ 4 } \right) = 0.1025}`

c) Find the probability if She gets the majority of the questions right.

We know that the probability mass function is :

`\[P\left( {X = x} \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\x\\\end{array}} \right){\left( {0.25} \right)^x}{\left( {1 - 0.25} \right)^(5 - x)}\]`

So; of She gets majority of her answers right ; we have:

The required probability is,

`P(X>2) = P(X=3) +P(X=4) + P(X=5)`

∴

`\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = \left( {\begin{array}{*{20}{c}}\\5\\\\3\\\end{array}} \right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^(5 - 3)}\] + \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)}\]`

`\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = (5!)/(3!(5-3)!)\right){\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^(5 - 3)}\] + (5!)/(4!(5-4)!) \right){\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^(5 - 4)}\] + (5!)/(5!(5-5)!) \right){\left( {0.25} \right)^5}{\left( {1 - 0.25} \right)^(5 - 5)}\]`

`\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = (5!)/(3!(5-3)!)\right){\left( {0.25} \right)^3}{\left( {1 - 0.75} \right)^(2)}\] + (5!)/(4!(5-4)!) \right){\left( {0.25} \right)^4}{\left( {0.75} \right)^(1)}\] + (5!)/(5!(5-5)!) \right){\left( {0.25} \right)^5}{\left( {0.75} \right)^(0)}\]`

`\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = 0.0879 + 0.0146 + 0.001`

`\mathbf{\[P\left( {X = 3 \ or \ 4 \ or \ 5 } \right) = 0.1035}`