Torpedoes Sinking A Ship

This problem can be viewd as writing words of length $4$ consisting only of the letters $s,d,m$ (for sink, damage and miss), and we want only the words with either at least one $s$ or at least two $d$ . Let's count them:

There are $3^4$ words in total, $1$ of which contains only $m$ and $4$ of which contain exactly $1$ $d$ and $3$ $m$ 's. So there are $3^4-1-4=81-5=76$ 'good' words out of $81$ .

Hence the probability to sink the ship is $\frac{76}{81}$ , and $a=76$ , $b=81$ . So $a+b=\boxed{157}$

The problem seems difficult because there are lots of ways to sink the ship; we can sink it on the first shot, we can damage it on the first shot, miss the second shot, and damage it again on the third shot, etc. So it should be much easier to solve for the probability of not sinking the ship (formally known as the complement). In other words, the probability that we do not sink the ship should equal 1 minus the probability that we do sink the ship.

Note that to avoid sinking the ship, we cannot ever fire a lethal shot, and we can only fire 0 or 1 damaging shots in the four shots that we get. We can easily use casework here for a clean solution:

Case 1: We fire no lethal shots and 0 damaging shots. The probability of case 1 is equivalent to missing 4 times, so the probability is: $P(miss 4 times) = (\frac{1}{3})^{4} = \frac{1}{81}$

Case 2: We fire no lethal shots and exactly 1 damaging shots. The probability of case 2 is the sum of the probabilities of the possible ways to damage with 1 shot and miss the other 3. There are four ways to do so: we can damage on either the first, second, third, or fourth shot. The probability of each of those four cases is 1/81, and since all four meet the criteria for case 2, the probability is $P(miss 3, damage 1) = 4(\frac{1}{81}) = \frac{4}{81}$

Thus we can add the probabilities of case 1 and case 2 to find the probability that we do not sink the ship in four shots: $P(don't sink) = \frac{1}{81} + \frac{4}{81} = \frac{5}{81}$

Now we can solve for the probability that we do sink the ship: $P(don't sink) = 1 - P(sink)$ $P(sink) = 1 - P(don't sink) = 1 - \frac{5}{81} = \frac{76}{81}$

So our final answer is 76 + 81 = 157 .

Lets use the complement. It can't either take one damaging hit, or not get hit for it not to be sunk. There is 4*(1/3)^4 probability to only get hit once, or 4/81. There is (1/3)^4 probability to not get hit, or 1/81. Add those to get 5/81. 1 - 5/81 = 76/81, so a = 76, b = 81. 76 + 81 = 157.

All miss: $(1/3)^{4}$ .
Then one damages and others all miss: $4\times (1/3)^{4}$
Required probability = $1-$ above 2 cases.

The ship will sink in case of a double blow. A single tornado can either give a double blow( sinking the ship) or a single blow( damaging the ship) or no blow at all( missing the ship). And since all the cases have equal probability, the generating function will be $( 1+x + x^{2} )^{4}$

And we have to find the sum of coefficients of all the terms with power greater than or equal to 2 ( ie no: of cases with two or more blows) =76.

And total number of cases is $3^{4}=81.$