A probability problem by Akshit Kumar

Given the information that he did get it right, he could have gotten it right in only one of these two ways:

1. If the student had known the answer (which has a $90\%$ probability), he would have definitely gotten the question right.

$\Longrightarrow P(right, knowledge) = 90\%$

1. If the student had not known the answer (which itself has a $10\%$ probability), he would have had $25\%$ chance of getting it right.

$\Longrightarrow P(right, guess) = 10\% \times 25\% = 2.5\%$

$\dfrac{a}{b} = \dfrac{2.5\%}{90\% + 2.5\%} = \dfrac{2.5\%}{92.5\%} = \dfrac{1}{37} \Longrightarrow 1+37 = \boxed{38}$

This question can be solved directly using the Bayes rule .

Bayes rule states that for two events $A$ and $B$ ,

$P ( A | B) = \dfrac{P(B | A) \times P(A)}{P(B)}$ where $P(A | B)$ denotes the conditional probability. (What is the probability of event $A$ occurring given that event $B$ has occurred)

Let $G$ denote the event that the student guesses the answer. Let $C$ denote the event that the answer is correct. Note that $\neg G$ denotes that the student is not guessing, or that he knows the answer.

We want to find the probability of the answer being guessed given that the answer is correct. In other words, we have to find $P (G | C)$

Now, $P ( G | C) = \dfrac{P(C | G) \times P(G)}{P (C)}$

• $P (C | G)$ , or the probability that the answer is correct given that it is guessed is simply $\dfrac{1}{4}$ since he randomly chooses one of the four answers.

• $P(G)$ is $1 - \dfrac{90}{100}$ since he knows the answer $\dfrac{90}{100}$ of the times. Therefore, he doesn't know the answer (and thereby has to guess it) is $\dfrac{10}{100}$ of the times.

• $P(C) = P(C | G) \times P(G) + P(C | \neg G) \times P(\neg G)$ . The probability that the answer is correct has two situations - One in which he guesses the answer, and one that he knows that answer. If he guesses the answer (which is $\dfrac{10}{100}$ ) of the times, the probability that he is correct is $\dfrac{1}{4}$ . If he knows that answer (which is $\dfrac{90}{100}$ ) of the times, the probability that he is correct is $1$

Filling in the values,

Now, $P ( G | C) = \dfrac{\dfrac{1}{4} \times \dfrac{10}{100}}{\dfrac{1}{4} \times \dfrac{10}{100} + 1 \times \dfrac{90}{100} }$

This equals $\dfrac{1}{37}$ , giving us the answer $\boxed{38}$

90% I know the solution and I get it right.

10% I don't know it and I try to guess. In this case I have 25% of guess it right, that is a 2.5% of the total possibilities.

Solution is: $\cfrac {2.5\%}{2.5\% + 90\% } = \cfrac {25}{25+900} = \cfrac{1}{37}$

$1+37 = \boxed{38}$

just use bayes theorem

The probability that the answer is correct if he knows the answer=0.9=9/10

The probability that the answer is correct if he doesn't know the answer = 0.1*1/4=1/40

Given that the answer is attempted correctly.

The probability that the answer is correct = 9/10 + 1/40

The probability that the answer is correct if he guesses it=(1/40)/( 9/10 + 1/40)=1/37

Hence, a=1 b=37

a+b=38

Defining the events : G->the event when the student guesses the answer. C->the answer is correct.

P(C)= 9/10 +1/10*1/4 = 37/40

P(G|C)= P(G intersection C)/ P(C) =(1/40)/(37/40)
=1/37

therefore; a=1 and b=37 so' a+b=38

First, we derive that the probability of the student getting the answer correct is $9/10 + 1/10 \times 1/4 = 37/40$ . This subset is what we are considering since we know that the student got the answer correct. Thus the probability that he was guessing is $\frac{ 1/10 \times 1/4 }{37/40} = \frac{1/40}{37/40} = \frac{1}{37} = \frac{a}{b}$ Thus $a+b=\boxed{38}$

$P(guessing|correct) = \frac{P(guessing \ and \ correct)}{P(correct)}$

$P(guessing \ and \ correct) = 0.1*0.25 = 0.025$

$P(correct) = 0.1*0.25 + 0.9*1 = 0.925$

$P(guessing|correct) = \frac{0.025}{0.925} = \frac{1}{37}$

$a + b = 1 + 37 = 38$

P(guess)= 1/10

P(correct|guess) = 1/4

P(correct)=9/10 + 1/40= 37/40

Using bayes theorem, we get (guess|correct) = 1/37

So our answer is 38