JEE-Mains

Let the probability that the student knows exactly $k$ answers be given by $\dfrac{m}{k}$ , where $m$ is a constant. Since the student must know a certain number of questions from $1$ to $90$ , it follows that $\sum \limits_{k=1}^{90} \dfrac{m}{k} = 1 \implies m \left( \sum \limits_{k=1}^{90} \dfrac{1}{k} \right)= 1 \implies m= \dfrac{1}{1/0.196751}= \dfrac{1}{0.196751}.$ We thus conclude the probability of the student knowing precisely $k$ questions is $0.196751 \times k$ . The expected number of questions the student knows is given by $\sum \limits_{k=1}^{90} k\dfrac{0.196751}{k} =90 \times 0.196751 = 17.70759.$ The expected number of questions the student doesn't know is $90-17.70759= 72.29241$ .

The expected marks the student gets from answering the questions he knows is $17.70759 \times 4$ . The expected number of questions the student guesses is $\dfrac{4}{5} \times 72.29241$ . Out of these guesses, the expected number of marks he scores is $72.29241\times \dfrac{4}{5} \times \dfrac{1}{4} \times 4 + \dfrac{4}{5} \times 72.29241 \times \dfrac{3}{4} \times (-1).$

Using Linearity of Expectation, we conclude the net expected score of the student is $17.70759 \times 4 + 72.29241\times \dfrac{4}{5} \times \dfrac{1}{4} \times 4 + \dfrac{4}{5} \times 72.29241 \times \dfrac{3}{4} \times (-1) = 85.28842.$ Since the question asks for the largest integer smaller than this, our desired answer is $\boxed{85}$ .

The probability of knowing exactly $k$ problems is directly proportional to $\frac{1}{k}$ and hence, can be taken as $\frac{p}{k}$ . Now, if $P_{k}$ denotes the probability of knowing exactly $k$ problems, then clearly,

$\displaystyle \sum_{k=1}^{90} P_{k} = 1$

$\Rightarrow p \displaystyle \sum_{k=1}^{90} \frac{1}{k} = 1$ $\Rightarrow p \frac{1}{0.196751} = 1 \Rightarrow p =0.196751.$

Now let us consider the event $E$ , that the student knows $90 - n$ problems, and does not know $n$ problems. Out of these $n$ problems , he attempts $k$ problems , and of these $k$ problems, $m$ get correct.

Total marks = $4(90 - n+m) + (-1)(k - m) = 360 - 4n+5m-k$

The probability that he knows $90 - n$ problems is $\frac{p}{90 - n}$ .

The probability that he marks $k$ problems out of $n$ is : ${n \choose k} (0.8)^k (0.2)^{n-k}$

The probability that out of $k$ problems, $m$ get correct is: ${k \choose m} (0.25)^m(0.75)^{k-m}$

Hence, the probability that event $E$ occurs is $\frac{p}{90 - n} \times {n \choose k} (0.8)^k (0.2)^{n-k} \times {k \choose m} (0.25)^m(0.75)^{k-m}$ .

The marks corresponding to event $E$ are $360 - 4n+5m-k$ .

Hence, the expected marks are :

$M = \displaystyle \sum_{n=0}^{89} \sum_{k = 0 }^{n} \sum_{m= 0}^{k} \bigg(\frac{p}{90 - n} \times {n \choose k} (0.8)^k (0.2)^{n-k} \times {k \choose m} (0.25)^m(0.75)^{k-m} (360 - 4n +5m - k)\bigg)$

To evaluate this summation, we use following property :

$\displaystyle \sum_{r=0}^{n} r {n \choose r} a^r (1-a)^{n-r}$

= $\displaystyle \sum_{r=0}^{n} na {n - 1 \choose r - 1} a^{r-1} (1-a)^{(n-1)-(r-1)}$ , by using $r {n \choose r} = n { n-1 \choose r-1}$

= $na (a+1-a)^{n-1} = na$

Hence, $\displaystyle \sum_{m=0}^{k} m {k \choose m} (0.25)^m(0.75)^{k-m} = \frac{k}{4}$ ,

Also, $(360 - 4n -k)\displaystyle \sum_{m=0}^{k} (0.25)^m(0.75)^{k-m} = 360 - 4n - k$

$\Rightarrow M = \displaystyle \sum_{n=0}^{89} \sum_{k = 0 }^{n} \bigg(\frac{p}{90 - n} \times {n \choose k} (0.8)^k (0.2)^{n-k} \bigg(360 - 4n + 5\frac{k}{4} - k \bigg)\bigg)$ ,

We use the same identity again to get:

$M = \displaystyle \sum_{n = 0}^{89} \frac{p}{90 - n} (360 - 4n +\frac{n}{4} \times 0.8)$

= $360p + \frac{p}{5} \displaystyle \sum_{n = 0}^{89} \frac{n}{90 - n}$

= $360p + \frac{p}{5}\bigg(90 \displaystyle \sum_{n=0}^{89} \frac{1}{90-n} - \sum_{n=0}^{89} 1\bigg)$

= $360 \times 0.196751 + \frac{0.196751}{5} \bigg(\frac{90}{0.196751} - 90\bigg) = \boxed{85.288}$

Note: Since the student knows $90-n$ problems, hence, $n$ varies from $0$ to $89$

Consider the probability with a fixed value of $k$ .

Our student (S) expected value can be calculated as follows:

Clearly he will get at least $k$ problems right, so he starts out with an expected value of $4k$ .

Also, since S attempts 80% of problems S doesn't know, S attempts $\frac{4}{5}(90 - k)$ additional problems in which he randomly guesses.

When randomly guessing, the expected value is $\frac{1}{4}(4) - \frac{3}{4}(1) = \frac{1}{4}$ marks.

Putting this together, the expected value given we know the value of $k$ is...

$4k + \frac{4}{5} (90 - k) (\frac{1}{4}) = \frac{19k + 90}{5}$

The probability that $k$ is a given value is $1/k \div 1/.196751 = \frac{.196751}{k}$ .

So, since k is not really given and is really a variable from 1 to 90, we have to factor in the probability that have that specific value of k into our previous expected value:

$\left( \frac{0.196751}{k} \right) \left( \frac{19k + 90}{5} \right)$

Our answer is the sum of the individual expected values for all possible values of $k$ :

$\sum_{k=1}^{90} \left( \frac{0.196751}{k} \right) \left( \frac{19k + 90}{5} \right)$

Now we simplify:

$(0.196751) \sum_{k=1}^{90} \frac{19k + 90}{5k}$

$(0.196751) \left\{ \sum_{k=1}^{90} 3.8 + \sum_{k=1}^{90} \frac{18}{k} \right\}$

$(0.196751) \left\{ \sum_{k=1}^{90} 3.8 + (18) \sum_{k=1}^{90} \frac{1}{k} \right\}$

$(0.196751) \left\{ (90)(3.8) + \frac{18}{0.196751} \right\}$

$0.196751 \cdot 90 \cdot 3.8 + 18 \approx \boxed{85}$