Pirate's Booty

Let choosing a gold coin be Event $A$ , let choosing nth basket be Event $B_n$ . Since $B_i$ are disjoint events, $P(A) = P(A|B_1) \times P(B_1)+ P(A|B_2)\times P(B_2) + P(A|B_3)\times P(B_3)$ , so $P(A) = (3/3)\times (1/3)+(2/4)\times (1/3)+(2/3)\times (1/3) = (13/18)$ . Thus, $a+b = 13+18 = 31$ .

Lets simple try to write the sample space of the given experiment-

S = {(1,G), (1,G), (1,G), (2,G), (2,G), (2,B), (2,B), (3,G), (3,G), (3,S)}

Here 1, 2, 3 refers to first, second and third casket respectively. And G, B, S refers to Gold, Bronze and Silver respectively.

Since all the sample points are not equally likely to happen the probabilities assigned to the 10 elementary events are - 1/9, 1/9, 1/9, 1/12, 1/12, 1/12, 1/12, 1/9, 1/9, 1/9,

Now the probability of getting a gold coin is P(G) = P(1,G) + P(1,G) + P(1,G) + P(2,G) + P(2,G) + P(3,G) + P(3,G) = 13/18 and hence follows the answer 13 + 18 = 41

There are 3 caskets to chose from. There is a 1/3 chance you chose the one with all three gold coins, if so the chance of drawing a gold coin is 1 multiplied by 1/3 yields 1/3. The chance that you chose the second casket is also 1/3 multiplied by the chance of drawing a gold coin (2/4 or 1/2) yields 1/6. Similarly the chance of choosing the third casket is 1/3 which is multiplied by the probability of drawing a gold coin from this casket (2/3) yields 2/9. Adding all of the above probabilities yields 13/18. Add these two integers yields 31.

There are 3 caskets each containing some coins . Therefore the probability that you choose one of the caskets = 1/3 .

Now the probability you choose a gold from the first casket = 1/3 * 3/3 since all are gold in the first casket = 1/3

The second one has 2 gold and 2 bronze thus probability of choosing a gold from the second casket is 1/3 *2/4 = 1/6

similarly from the third casket the probability is 2/9 .

Add the three solutions = 1/3 +1/6 + 2/9 , this is the probability that you pick a gold . the solution = 13/18 (a/b) . Question is sum (a,b) = 31

There is an equal chance of drawing a coin from any of the three caskets - 1/3 for each. Within that 1/3, however, the chance of drawing a gold coin is further limited.

The first casket is contains only gold coins. Hence, there is a 100% chance of drawing a gold coin from the first casket, giving a total of a 1/3 chance (1/3 * 1/1).

The second casket contains 2 gold coins and 2 bronze coins. Hence, there is a 50% chance of drawing a gold coin from it, giving a total of a 1/6 chance (1/3 * 1/2).

The final casket contains 2 gold coins and 1 silver coin. Hence, there is a 2/3 chance of drawing a gold coin from it, giving a total of a 2/9 chance (1/3 * 2/3).

The final probability consists of adding up all these chances (1/3 + 1/6 + 2/9), giving us a final total of 13/18. All that remains is to sum 13 and 18, giving 31.

While the solution is simple, you have to look at each individual case by itself. There are three caskets so the probability of drawing gold must be broken up into those three separate cases.

Casket 1 contains $3$ gold coins, so the probability of drawing gold is equal to $1$ . That said, the probability of choosing the first casket (or any, for that matter) is $\frac {1}{3}$ so multiplying those two numbers together, you would get the combined probability of drawing gold from the first casket to be $\frac {1}{3}$

Casket 2 contains $2$ gold and $2$ bronze, so a $\frac {2}{4}$ probability of drawing gold can be simplified to $\frac {1}{2}$ . Again, the probability of choosing any casket is equal to $\frac {1}{3}$ , so when multiplying the numbers together, the probability of drawing gold from the second casket is $\frac {1}{6}$ .

Casket 3 , following pattern, contains $2$ gold and $1$ silver leaving us with a $\frac {2}{3}$ probability of drawing gold. With a $\frac {1}{3}$ chance of picking the third casket, we multiply those two numbers together for a combined probability of $\frac {2}{9}$ of drawing gold.

Finally, we add these three probabilities together. $\frac {1}{3} + \frac {1}{6} + \frac {2}{9}$ = $\frac {13}{18}$ . From this we can conclude $a=13$ and $b=18$ . $a + b$ must therefore be $31$ .

the probability to choose a casket is 1/3. In the first casket, the probability of getting a gold coin is 1. In the second, 1/2. And in the third is 2/3. But like I said earlier, the probability to choose a casket is 1/3, so lets multiply the probabilities of getting a gold coin in a casket to by 1/3. the over-all probability of getting a gold coin is (1 x 1/3)+(1/2 x 1/3)+(2/3 x 1/3) which is equal to 1/3 + 1/6 + 2/9 which results to 13/18 where a=13 and b=18 therefore, a+b=13+18=31

let a/b be the probability of drawing a gold coin

first: the probability that a casket is chosen is 1/3.

second: the probability of choosing a gold in each casket are as follows

         Casket 1 = 1 (or 100%), Casket 2 = 1/2, and Casket 3 = 2/3

third: we multiply the probability of choosing a casket to individual probability of choosing a coin in each casket

        i.e. 1/3*( 1 + 1/2  + 2/3) = 13/18

fourth: we add 13 and 18 since it represents a and b.

       i.e. 13 + 18 = 31

The probability of choosing each casket is $\frac{1}{3}$ .

There is a $\frac{3}{3}$ chance of choosing a gold coin from the first casket so the probability of choosing the first casket and a gold coin from that same casket is $\frac{1}{3}$ .

There is a $\frac{2}{4}$ chance of choosing a gold coin from the second casket so the probability of choosing the second casket and a gold coin from that same casket is $\frac{1}{6}$ .

There is a $\frac{2}{3}$ chance of choosing a gold coin from the third casket so the probability of choosing the third casket and a gold coin from that same casket is $\frac{2}{9}$ .

The sum of these three probabilities is $\frac{13}{18}$ .

$13+18=31$

 The probability of gold from FIRST CASKET is 3/3.  The probability of gold from SECOND CASKET is 2/4.  The probability of gold from THIRD CASKET is 2/3.  So the required probability is 1/3 (3/3+2/4+2/3)=13/18=a/b  a+b= 31

selection of any casket is 1/3.Hence the required probability is(1/3)(1+1/2+2/3) =13/18.where 1=probability of getting gold coin from the first casket,1/2 is from second and 2/3 is from 3rd.hence a+b=13+18=31.

The probability of drawing from any individual casket is $\frac {1}{3}$ . For the first casket, the probability of drawing a gold coin is 1. For the second casket, the probability of drawing a gold coin is $\frac {1} {2}$ . For the third casket, the probability of drawing a gold coin is $\frac {2}{3}$ .

Hence, the probability that you drew a coin coin is

$\begin{aligned} & P( \text{coin from first casket is gold}) \times P ( \text{drawn from first casket} ) \\ &+ P( \text{coin from second casket is gold}) \times P( \text{drawn from second casket} ) \\ &+ P( \text{coin from third casket is gold}) \times P \text{drawn from third casket} ) \\ &= \frac {1}{3} \cdot 1 + \frac {1}{3} \cdot \frac {1}{2} + \frac {1}{3} \cdot \frac {2}{3} \\ &= \frac {13}{18} \\ \end{aligned}$

Hence, $a + b = 13 + 18 = 31$ .

Note: Why can't we say that out of 10 coins, there are 7 gold coins, so the probability is $\frac {7}{10}$ ?