Run & Hide

Label each friend 1,2,3, and 4.

If a ghost opens up a door, the probability of not seeing friend 1 is 3/4.

The probability of not seeing friend 2 is 3/4.

The probability of not seeing friend 3 is 3/4.

The probability of not seeing friend 4 is 3/4.

The probability of all these events happening at the same time is (3/4)^4.

3^4+4^4=337

Relevant wiki: Conditional Probability Distribution

The total combinations for these $4$ teenagers to hide in the $4$ rooms = $4^4 =256$ ways.

First, if all of them decide to hide in one single room, there are obviously $\binom{4}{1} = 4$ ways to choose a room. Should this happen, the ghost will have a probability of $\dfrac{3}{4}$ to find no-one when opening one room randomly.

Second, if two rooms are occupied by these teenagers, the combinations will equal to $\binom{4}{2}[\left\{4 \atop 2\right\}(2!)] = 12(\binom{4}{3} + 3\binom{4}{4}) = 84$ ways. In this case, the probability of finding no-one = $\dfrac{2}{4}$ .

Third, if three rooms are occupied by these teenagers, the combinations will equal to $\binom{4}{3}[\left\{4 \atop 3\right\}(3!)] = 4\times 6\times \binom{4}{2} = 144$ ways. In this case, the probability of finding no-one = $\dfrac{1}{4}$ .

Finally, if all the four rooms are occupied, there will be $4! = 24$ ways for each teenager to independently hide in. However, in this case, the probability of the ghost's finding no-one is $0$ ; there will always be someone in any room.

Hence, the probability of finding no-one in the room = $\dfrac{(4\times \dfrac{3}{4}) + (84\times \dfrac{2}{4}) + (144\times \dfrac{1}{4}) + (24\times 0) }{256} = \dfrac{3 + 42 + 36}{256} = \dfrac{81}{256}$ .

Alternatively, we can rethink of the scenario as one of the four rooms is the death door, which the ghost will choose. Then the events of the 4 teenagers will choose the death door are independent from one another, and the probability of each not seeing the ghost = $\dfrac{3}{4}$ .

Hence, the probability that all will not see the ghost = $(\dfrac{3}{4})^4 = \dfrac{81}{256}$ .

Thus, $a+b = 337$ .

Let us denote the probability that i th friend is caught by P( $A_{i}$ )

The probability of each friend getting caught is same and equal to:

P( $A_{1}$ ) = P( $A_{2}$ ) = P( $A_{3}$ ) =P( $A_{4}$ ) = $\frac{1}{4}$

Probability that ith friend is not caught =P( $\overline{A_i}$ )= 1-P( $A_{i}$ ) = $\frac{3}{4}$

The 4 events ( of each friend getting caught) are independent events

Thus, Probability that no friend is caught is:

P( $\overline{A_1}$ $\bigcap$ $\overline{A_2}$ $\bigcap$ $\overline{A_3}$ $\bigcap$ $\overline{A_4}$ ) = P( $\overline{A_1}$ ) $\times$ P( $\overline{A_2}$ ) $\times$ P( $\overline{A_3}$ ) $\times$ P( $\overline{A_4}$ )

= $\frac{3}{4}$ $\times$ $\frac{3}{4}$ $\times$ $\frac{3}{4}$ $\times$ $\frac{3}{4}$

= $\frac{81}{256}$

Assume WLOG the monster will always pick room A. Then, for the gang to not get caught they would need to hide in rooms B, C, or D to not get caught. With this, each member has a 3/4 chance of being safe, or $(3/4)^4$ chance the whole gang is safe. Therefore, it's a 81/256 chance, and gives 81+256=377.

Suppose the monster is going to open a door, what is the probability that a friend is not hiding in there? 3/4. Then we have just to raise it to the fourth power