There are three boxes: one with two golden coins, one with two silver coins, and one with one golden coin and one silver coin. After picking one of the boxes with equal probability, and then one coin from the selected box, it turns out that it is a silver coin. What is the probability that the other coin in the chosen box is also silver

Step-by-step explanation:

From Bayes' theorem is stated mathematically as the following equation:[2]

{\displaystyle P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},}

where A and B are events and P(B) ≠ 0.

P(A) and P(B) are the probabilities of observing A and B without regard to each other.

P(A | B), a conditional probability, is the probability of observing event A given that B is true.

P(B | A) is the probability of observing event B given that A is true.

At this point, go through the attached file before you continue with part B.

Part B)

P(silver) = P(silver from SS)+P(silver from GS)

note P(SS)=P(GG)=P(GS) = 1/3

P(silver from SS) = 1

P(silver from GS) = 1/2

hence

P(Silver from SS) = 1/3

P(Silver from GS) = 1/3 *1/2

P(Silver) = 1/3*1+1/3*1/2

required probability = P(Silver from SS)/P(Silver) = 2/3

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

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