The Absent-Minded Professor

We know the professor left his umbrella in a room so we need to calculate the probability he left his umbrella for each of the rooms.

The probability the professor left his umbrella in the first room is $\frac{1}{4}$ .

The probability the professor left his umbrella in the second room is the probability he didn't leave it in the first room multiplied by a fourth which is $\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{16}$ .

The probability the professor left his umbrella in the third room is the probability he didn't leave it in the first two rooms multiplied by a fourth which is $\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{1}{4}=\frac{9}{64}$ .

Now we want to find the probability he left his umbrella in the first room given he left his umbrella. The probability he left his umbrella is $\frac{1}{4}+\frac{3}{16}+\frac{9}{64}=\frac{37}{64}$ . The probability he left his umbrella in the first room is $\frac{1}{4}$ . Thus the probability he left his umbrella in the first room given he left his umbrella is $\frac{1}{4}/\frac{37}{64}=\frac{16}{37}$ . $16+37=53$ .

Let $A$ be the event that the umbrella is lost in the first classroom and $B$ be the event that the umbrella is lost at all. Then, by the law of conditional probability $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$.

Since $A$ only occurs when $B$ occurs, we can rewrite the equation as $$P(A|B) = \frac{P(A)}{P(B)}$$.

We are given that $P(A) = \frac{1}{4}$ , so we only need to find $P(B)$ . It is the sum of probabilities that the professor loses the umbrella in each room, therefore $$P(B) = \frac{1}{4} + \frac{3}{4}\cdot\frac{1}{4} + \frac{3}{4}\cdot \frac{3}{4}\cdot\frac{1}{4} = \frac{37}{64}$$.

From here, we can calculate $P(A|B) = \frac{1}{4} / \frac{37}{64} = \frac{16}{37}$ . Therefore, the answer is $\boxed{53}$ .

Let A $\equiv$ Event of forgetting umbrella in first classroom and B $\equiv$ Event of forgetting umbrella in any classroom By Bayes Theorem $P(A | B)*P(B) = P(A \cap B)$ $But A \subset B$ $So P(A | B) = \frac{P(A)}{P(B)}$ $\Rightarrow P (A | B) = \frac{\frac{1}{4}}{1-\frac{3*3*3}{4*4*4}}$ $\Rightarrow P (A | B) = \frac{\frac{1}{4}}{\frac{37}{64}}$ $\Rightarrow P (A | B) = \frac{16}{37}$

So answer is 16 + 37 = 53

LetA be the event that the umbrella is lost in the first classroom and be the event that the umbrella is lost at all. Then, by the law of conditional probability $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$. Since only occurs when occurs, we can rewrite the equation as $$P(A|B) = \frac{P(A)}{P(B)}$$. We are given that , so we only need to find . It is the sum of probabilities that the professor loses the umbrella in each room, therefore $$P(B) = \frac{1}{4} + \frac{3}{4}\cdot\frac{1}{4} + \frac{3}{4}\cdot \frac{3}{4}\cdot\frac{1}{4} = \frac{37}{64}$$. From here, we can calculate . Therefore, the answer is

This is a simple application of Bayes' theorem. An event, namely the professor leaving his umbrella in one of the classrooms, has already happened.Lets call this event B.

We are asked to compute the probability that he left the umbrella in the first classroom, given that he returned to office without the umbrella. Let's call the event that he left the umbrella in the first classroom A. Then, effectively we have to find the probability of event A given B has already happened. Mathematically, we need to find P( A|B).

Now, we can use Bayes' theorem here:

P(A|B) =( P(A)*P(B|A) ) / P(B)

Now:

• P(A) is given in the question, i.e. 1/4.
• P(B|A) = 1, since B will always happen if A has happened.
• P(B) = probability he forgets umbrella in first classroom + probability he doesn't forget umbrella in first classroom * probability he forgets umbrella in second classroom + probability he doesn't forget umbrella in first classroom * probability he doesn't forget umbrella in second classroom * probability he forgets umbrella in third classroom = $1/4 + 3/4 \times 1/4 + 3/4 \times 3/4 \times 1/4$ = 37/64.

Thus, P(A|B) = ((1/4*1)/(37/64)) = 16/37.

Thus, a = 16 and b=37 which gives a+b = 53