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In an examination, a student can choose the order in which two questions ($\textsf{QuesA}$ and $\textsf{QuesB}$) must be attempted.

• If the first question is answered wrong, the student gets zero marks.
• If the first question is answered correctly and the second question is not answered correctly, the student gets the marks only for the first question.
• If both the questions are answered correctly, the student gets the sum of the marks of the two questions.

The following table shows the probability of correctly answering a question and the marks of the question respectively.

$$\begin{array}{c|c|c} \text{question} & \text{probabilty of answering correctly} & \text{marks} \\ \hline \textsf{QuesA} & 0.8 & 10 \\ \textsf{QuesB} & 0.5 & 20 \end{array}$$

Assuming that the student always wants to maximize her expected marks in the examination, in which order should she attempt  the questions and what is the expected marks for that order (assume that the questions are independent)?

1. First $\textsf{QuesA}$ and then $\textsf{QuesB}$. Expected marks $14$.
2. First $\textsf{QuesB}$ and then $\textsf{QuesA}$. Expected marks $14$.
3. First $\textsf{QuesB}$ and then $\textsf{QuesA}$. Expected marks $22$.
4. First $\textsf{QuesA}$ and then $\textsf{QuesB}$. Expected marks $16$.

### 1 comment

Expected marks from question 1 = $0.8\times10 = 8$

Expected marks from question 2 = $0.8 \times 0.5 \times 20 = 8$

Expected total marks = 16, I don’t get how the order in which she attempts matter but I marked option D.

First we answer A, then B:

• $\text{Expected marks}= \text{Probability}_{\text{A is wrong}} *0+ \text{Probability}_{\text{A is correct}} * \text{Probability}_{\text{B is wrong}}*10+ \text{Probability}_{\text{A is correct}}* \text{Probability}_{\text{B is correct}}*30$
$=0.2*0+0.8*0.5*10+0.8*0.5*30$
$=0+4+12$
$=16$

First we answer B, then A:

• $\text{Expected marks}= \text{Probability}_{\text{B is wrong}} *0+ \text{Probability}_{\text{B is correct}} * \text{Probability}_{\text{A is wrong}}*20+ \text{Probability}_{\text{B is correct}}* \text{Probability}_{\text{A is correct}}*30$
$=0.5*0+0.2*0.5*20+0.8*0.5*30$
$=0+2+12$
$=14$

So to maximize marks we should first answer $A$ and then $B.$ The expected marks in such a scenarion will be $16.$

Answer is $D$.

$\textbf{A then B -}$

A B Expected Marks
Correct Correct

30

Correct Wrong 10
Wrong Correct 0
Correct Wrong 0

Expected Marks = $(0.8)*(0.5)*30 + (0.8)*(0.5)*10 = 16$

$\textbf{B then A -}$

B A Expected Marks
Correct Correct

30

Correct Wrong 20
Wrong Correct 0
Correct Wrong 0

Expected Marks = $(0.5)*(0.8)*30 + (0.5)*(0.2)*20 = 14$

Attempt $A$ then $B$ and get expected marks $16$.

### 1 comment

Please correct the last row in both the tables.

 A B Correct Correct Correct Wrong Wrong Correct Wrong Wrong

Attempt QuesA first and then QuesB,

Expected Marks = $0.8*0.5*10 + 0.8*0.5*30 = 4 + 12= 16$

Attempt OuesB first and then QuesA,

Expected marks = $0.5*0.2*20 + 0.5*0.8*30 = 2 + 12 = 14$

Option D is correct.

We can see here 2 cases,

1. If she attempt 1st Qa , then Qb
1. If she attempt 1st Qb and then Qa

Qa = Attempt right , Qa’ = incorrect attempt ;

Qb =  Attempt right, Qb’ = incorrect attempt

Now we see this two cases by tree method ----

Case 1:

Marks for both correct = (10 + 20) = 30

Only for Qa correct = 10 marks

So, Expected marks = (0.8 x 0.5 x 30) + (0.8 x 0.5 x 10) = 16

Case 2:

Marks for both correct = (10 + 20) = 30

Only for Qb correct = 20 marks

So, Expected marks = (0.8 x 0.5 x 30) + (0.5 x 0.2 x 20) = 14

So, we can make this conclusion that,  Case 1 expected marks > Case 2 expected marks

So, She should do first Qa and then Qb Expected marks 16.

Hence, option D is correct

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