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.

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12 votes

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)?

- First $\textsf{QuesA}$ and then $\textsf{QuesB}$. Expected marks $14$.
- First $\textsf{QuesB}$ and then $\textsf{QuesA}$. Expected marks $14$.
- First $\textsf{QuesB}$ and then $\textsf{QuesA}$. Expected marks $22$.
- First $\textsf{QuesA}$ and then $\textsf{QuesB}$. Expected marks $16$.

14 votes

Best answer

**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 $16.$

5 votes

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 $16$.

0 votes

We can see here 2 cases,

**If she attempt 1st Qa , then Qb**

**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 *