Coins in Caskets

Let: the coins in casket 1 to be A and B ( A = gold , B = gold ), the coins in casket 2 to be C and D ( C = gold, D = silver ), the coins in casket 3 to be E and F ( E = silver, F = silver ). Therefore, the possible outcomes : Case 1: A (1st pick) B (2nd pick)
Case 2: B (1st pick) A (2nd pick)
Case 3: C (1st pick) D (2nd pick)
Case 4: D (1st pick) C (2nd pick)
Case 5: E (1st pick) F (2nd pick)
Case 6: F (1st pick) E (2nd pick)

Note that A, B and C to be gold coins The 1st coin picked is gold, the probability of getting 2nd coin that is gold is 2/3 ( Case 1, Case 2, Case 3 ) Therefore, 2 + 3 = 5

Let A and B be two events. Then, we say that $p(A|B) = \frac{p(A \cap B)}{p(B)}$ The expression on the left side means "the probability of A given B". If we take the gold coin from Casket 2, there is no way there is another one in it afterwards. Also, we cannot take any gold coin from Casket 3. Thus, the only possibility is getting one gold coin from Casket 1. The probability of taking a gold coin from Casket 1 is 1/3. The probability of taking any gold coin is 1/3 + 1/6 + 0. Applying the Conditional Probability formula for the events of "taking a gold coin from Casket 1" and "took a gold coin" (from any casket), we may say: $p(A|B) = \frac{1/3}{1/2} = \frac{2}{3}$ Which gives the answer to this question.

Let event G1 be 1st coin is gold, G2 be 2nd coin is gold. P( G2 if G1) = P(G1 and G2) / P(G1)

Event (G1 and G2) will happen only if there are 2 gold coins, that is the first casket is chosen , which has probability of 1/3.

Let choice be (C,c) where C is casket and c is coin.

Event (G1) = (1,G) + (2,G) +(3,G) P(G1) = P(1,G) + P(2,G) + P(3,G) This is equal to (1/3)(1) + (1/3)(1/2) + (1/3)(0) = 1/2

P( G2 if G1) = (1/3) / (1/2) = 2/3 As 2, 3 are coprime positive integers, a+b =5.

Ignoring the baskets, there are six possible coins for you to take first, three gold and three silver. Since we know the first coin was gold, we can ignore three of the six original possibilities.

Out of the current three possible coins for you to have chosen first, two were in the basket with two gold coins, therefore the probability that the next coin in the basket will be gold (a/b) is 2/3 and the sum a+b is 2+3 is 5.

Let , G be the event that the selected coin is gold.

Let the probability of selecting a gold coin from : 1st casket = P(G|C 1) = \frac{2}{2+0} = 1 2nd casket = P(G|C 2) = \frac{1}{1+1} = \frac{1}{2} 3rd casket = P(G|C_3) = \frac{0}{0+2} = 0

Probability of selecting each casket is P(C 1)=P(C 2)=P(C_3)=\frac{1}{3}

From bayes' theorem we have , the probability of 1st casket being selected given that the selected coin is gold is P(C 1|G) = \frac{P(G|C 1) P(C_1)}{P(G|C_1) P(C 1)+P(G|C 2) P(C_2)+P(G|C_3) P(C_3)} = \frac{1}{\frac{3}{2}}=\frac{2}{3}

similarly , probability of 2nd casket being selected is P(C 2|G) = \frac{P(G|C 2) P(C_2)}{P(G|C_1) P(C 1)+P(G|C 2) P(C_2)+P(G|C_3) P(C_3)} = \frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{3}

and probability of 3rd casket being selected is P(C 3|G) = \frac{P(G|C 3) P(C_3)}{P(G|C_1) P(C 1)+P(G|C 2) P(C_2)+P(G|C_3) P(C_3)} = \frac{0}{\frac{3}{2}}=0

These three possibilities are disjoint.

First casket have 2 gold coins. So if the selected gold coin comes from 1st casket then the other coin will be gold with probability 1. If the selected gold coin comes from 2nd casket then the other coin will be gold with probability 0 (as there are no more gold coins) and if the selected gold coin comes from 3rd casket then the other coin will be gold with probability 0 (the selected gold coin can't be from this casket because it has no gold coins).

So, now we have the probabilities of a casket being selected if the fist coin selected is gold. So the other selected coin will be gold with probability = (\frac{2}{3} 1)+(\frac{1}{3} 0)+(0*0) = \frac{2}{3}

If we think it as form \frac{a}{b} , then a=2 & b=3. so a+b=5

Let $X$ be the event of choosing gold as the first coin and let $Y$ be the event of choosing gold as the second coin. So, we look for the conditional probability $P(Y | X) = \frac{P (Y \cap X)}{P(X)}$ .

We have $P(X) = \frac{1}{3}\times 1 + \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 0 = \frac{1}{2}$ and $P(Y \cap X) = \frac{1}{3}$ as Casket 1 is the only casket that has $2$ gold coins.

Thus $P(Y | X) = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$ . Hence $a + b = 2 + 3 = 5$ .

CASKET 1 2 3

no./type (coins) 2G; 1G,1S; 2S

Probability of drawing gold coins (1st time) = no. of gold coins / total no. of coins = 3/6 = 1/2

Since 1st coin is G, therefore casket must be no. 1 or 2 and 2nd coin must be G or S.

Probability that 2nd time coin is G = 1 / 2

Therefore, probability that 2nd coin is gold = 1/2 * 1/2 = 1/4 = a/b

hence, a+b = 5

there are only two possible caskets that is being drawn given that the first coin is gold. therefore, the second coin can either be gold if it is casket 1 or silver if it is casket 2. since there are 3 golds and 1 silver in the two caskets a 1 is being drawn, there will only be 2 gold coins and i silver coin which means that the probability of getting a gold coin is 2 out of 3. and the sum of it is FIVE.

Consider only Casket 1 and Casket 2 so there are 3 gold in all and you need 2 therefore probability of gold = 2/3 or 2+3 = 5 :)

Therefore you will add the probability of first casket and the second casket 2/2+1/2 = 3/2 = a/b a=3 b=2 a+b=2+3=5

The first coin that was picked was gold, and there are only 2 casket that contains gold coin. Since there are only 2 casket to chose from the 3, a/b will be 2/3. 2/3 are coprime numbers since both can only be divided by one. 2+3 = 5

If there are three gold and three silver coins among the three caskets "to" serious gold and would have value 3, so "b" and would also would be worth three silver coins. if one of gold is removed reduces the possibilities to 2/3 then would be the sum of a + b = 2 +3 = 5