Gold and Silver Coins

Intuitive Explanation: There are two ways to get a gold coin: from chest A (from which we have 100% chance), or from chest B (from which we have 50% chance). Since we are twice as likely to get a gold coin if we are choosing from chest A, the odds that we chose from chest A are $2:1,$ so the probability is $\frac{2}{3}.$

Formal Explanation: We can apply Bayes' Theorem to this problem. Let $G$ be the event of selecting a gold coin, and note that $P(G|A) = 1$ and $P(G|B) = \frac{1}{2}.$ Via Bayes' Theorem, $P(A|G) = \frac{P(A)\cdot P(G|A)}{P(G)} = \frac{P(A)\cdot P(G|A)}{P(A)\cdot P(G|A) + P(B)\cdot P(G|B)} = \frac{\frac{1}{2} \cdot 1}{\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2}} = \frac{2}{3}.$

When we choose randomly a gold coin, from two boxes of 100 gold, 50 gold and 50 silver, total out come is 3 and favourable out come is two. Therefore Probabilit of getting gold coin is 2/3.

There are 150 gold coins in both chests combined. Chest A has twice as many as Chest B. Knowing that you chose a gold coin, look at your entire universe of possibilities as 150 coins. 100 of those lie in Chest A (100/150), making the probability that you chose from Chest A: 2/3.

The probability of getting a gold coin in chess A is 1/2. The probability of getting a gold coin in chess B is 1/2 x 1/2 = 1/4. If the coin you choose is gold, the probability that you chose chest A = 1/2 / (1/2 + 1/4) = 2/3

We need to find P(A|G)-ie the probability of picking up from A given evidence is -it is a Gold coin. P(G|A): prob of gold if chest is A=1. P(A): P(A)=P(B)=0.5. P(G)=150/200 =3/4. From Bayes P(A/G)=P(G/A) * P(A)/P(G)= 1 * 0.5 * 4/3= 2/3

P(H) = 0.5 P(E) = 3/4

P(E/H) = 1

applying baye's formula 2/3

Probability to extract gold in total is P(gold)=150/200 = 3/4; Probability to open chest A is P(A) = 1/2 that's the prior ; probability to extract gold form chest A is P(gold|A) = 1; Plugging in together in the formula: P(A|gold)=(P(gold|A)/P(gold)) P(A) = (1/(1/2)) 3/4=2/3

There are 150 gold coins. 100 are in Chest 1. 50 are in Chest 2. The probability the gold coin comes from chest 1 is 100/150 or 2/3.

Yayyy, i got right :D lol well I'm just trying to put my understanding in consideration from the Bayes Theorem {may not be accurate explanation}

Probability of choosing the Treasure A out of Two Treasures is 1/2 aka Prior Probability

Probability of Choosing the Gold Coin out of total available coins is 150 / 200 = 3/4

Probability of Choosing the Gold Coin exclusively from the Treasure A is 100/100 = 1 {It is so obvious that there are cent percent chances}

Probability for choosing the Treasure A given the evidence that the coin is Gold = Prior Probability Times likelihood ratio = (1 / 2) * ( 1 / (3/4)) = (1/2) * (4/3) = 2/3

X -- Chance to Select chest A Y -- Chance to select Gold coin Need to find p(X|Y) ? p(X|Y) = (p(X) p(Y|X))/p(Y) p(X) = 1/2 , p(Y) = 150/200 = 3/4 , p(Y|X)=1 p(X|Y) =(1/2 1)/3/4 = 2/3

• Using a more intuitive approach, in chest A there are 100 gold coins and in chest B there are 50 gold coins so the ratio of gold coins A : B = 2 : 1, hence the probability a person chooses A from a (virtual) chest that contains both A and B is 2 / (2 + 1) = 2/3

Known : - P(A) - probability of selecting chest A = 1/2, - P(G) - probability of selecting gold coin = 150/200 = 3/4, - P(G|A) - probability that the coin is gold given you selected from chest A = 1 (as all the coins are gold in chest A),

To find : P(A|G) - probability that the coin is from chest A given the coin is a gold coin, P(A|G) = $\frac{P(G|A) . P(A)}{P(G)}$ = $\frac{2}{3}$

P(A|G) = P(G|A) * P(A) / P(G) = 1 * (1/2) / (3/4) = 2/3

p(a/gold)=p(a) p(gold/a) / p(gold) =2/3