Chapter 12, Problem 15STP

a)

To determine

To find : Probability of getting correct answer in the test.

Answer to Problem 15STP

  $\frac{1}{4}$

Explanation of Solution

Given Information :

We know that, $(a,b,c)$ A test contains $8$ MCQ. A student answered by guessing.

Formula Used:

Binomial distribution, mean and variance. {for all $a,b,c$ }

Calculation :

There are $4$ choices in each question of which only one is correct.

  $\therefore$ Probability of getting one correct answer is given by

  $P=\frac{1}{4}$

b)

To determine

To find : On attempting each question, what is the expected number of correct answer ?

Answer to Problem 15STP

  $2$

Explanation of Solution

Given Information :

  $\text{(a, b, c) A test contains 8 MCQ}\text{. A student answered by guessing}\text{.}$

Formula Used:

Binomial distribution, mean and variance. {for all a, b, c}

Calculation :

  $\begin{array}{l}\text{Let X denote the number of correct answers, then according to}\\ \text{binomial distribution, n = 8, p = 1/4, q = 3/4}\\ \text{Expection of X is}\mu =np=2\end{array}$

c)

To determine

To find : Probability of getting atleast half correct answers.

Answer to Problem 15STP

  $\frac{933}{8192}$

Explanation of Solution

Given Information :

  $\text{(a, b, c) A test contains 8 MCQ}\text{. A student answered by guessing}\text{.}$

Formula Used:

Binomial distribution, mean and variance. {for all a, b, c}

Calculation :

  $\begin{array}{l}P(x\ge 4)=P(4)+P(5)+P(6)+P(7)+P(8)\\ =1-\left\{P(0)+P(1)+P(2)+P(3)\right\}\\ =1-\left\{{}^8{C}_0{\left(\frac{3}{4}\right)}^8+{}^8{C}_1\cdot \frac{1}{4}\cdot {\left(\frac{3}{4}\right)}^7+{}^8{C}_2{\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^6+{}^8{C}_3{\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^5\right\}\\ =1-\left\{\frac{6561+17496+20412+13603}{65536}\right\}\\ =\frac{7464}{65536}=\frac{933}{8192}\end{array}$