Alien Drugs 1

One jar will contain the orange candy and some number (between 0 and 9) of the greens (we'll call this the bad jar); the remaining greens will be distributed across other jars ("good" jars). Naturally we wish to minimize the odds of choosing the bad jar, which is accomplished by placing one green in each good jar.

Let $N$ be the number of nonempty jars in such a distribution. There is 1 bad jar and $N - 1$ good jars. Each good jar contains 1 green candy, and the bad jar contains 1 orange candy and $9 - (N - 1) = 10 - N$ green candies for a total of $11 - N$ candies. Thus the odds of choosing the orange candy become $(1/N)(1/(11 - N))$ , and the odds of survival are $1 - 1/(N(11 - N))$ . This quantity is maximized when $N(11 - N)$ is maximized, which is when $N$ is 5 or 6, for which the probability is $29/30$ , so $A + B = 59$ .

Obviously, the best strategy would be to put singular green candies in some jars, and multiple green ones in the jar containing the orange candy.

We first find out the probability for all the jars singularly filled with a candy.

P(Survival) = 9/10

Then we put one more green candy in the orange candy jar,

P(Survival) = 8/9 + (1/9)×(1/2)=17/18

Which is better.

We also notice that P(Survival) for all the candies in one jar is also 9/10, thus the P first increased and then decreased back to the same value, which gives us a hint that the max probability would be when half (nearly half) of the green candies are in jars (singularly), and the other half in the orange candy jar. Thus,

P(Max survival) = 5/6 + (1/6)×(4/5) = 29/30.

Let the probability of being poisoned be $q$ . Then the probability of surviving is $p=1-q$ . Now considering putting one candy in each of the 10 jars, then $q=\frac 1{10}$ . If we remove a jar and put a green candy with the poisonous orange candy, then $q = \frac 19 \times \frac 12 = \frac 1{18}$ , where $\frac 19$ is the probability of choosing the jar containing the green and orange candies and $\frac 12$ is the probability of choosing the orange candy from the two candies. We note that $q$ is greatly reduce. If now we remove one more jar and put two green candies with the orange candy, then $q = \frac 18 \times \frac 13 = \frac 1{24}$ . Again $q$ is reduced.

If we let the number of jar be $n$ , then the mortality probability is given by:

$\begin{aligned} q_n & = \frac 1n \times \frac 1{11-n} = \frac 1{n(11-n)} = \frac 1{11} \left(\frac 1n + \frac 1{11-n} \right) \end{aligned}$

By AM-GM inequality $q_n \ge \frac 2{11\sqrt{n(11-n)}}$ and equality occurs or $q_n$ is minimum when $n=5.5$ . For integer $n$ , minimum mortality probability is $q_5 = q_6 = \frac 1{30}$ and the maximum survival probability is $p = 1 - \frac 1{30} = \frac {29}{30}$ . Therefore, $A+B= 29+30 = \boxed{59}$ .