Calculating Probability of Coin Selection Using Bayes Rule

[ifly2hi]

Homework Statement

I am just wanting to get a good starting point... not an answer to the question. Question: A desk contains three drawers. Drawer 1 has two gold coins. Drawer 3 has one gold coin and one silver coin. Drawer 3 has two silver coins. I randomly choose a drawer and then randomly choose a coin. If a silver coin is selected, what is the probability that i choose drawer 3? I am suppose to use Bayes rule to find probability. Thank you for any help

Homework Equations

Bayes Rule that I used: P(S/D3)= [P(D3/S)P(D3)] / [P(D1/S)P(S) + P(D2/S)P(S)+P(D3/S)P(S)]

The Attempt at a Solution

I defined all I could: Drawer(D), Silver(S) and Gold(G).
P(D1)=P(D2)=P(D3)= 1/3 b/c he has a one in three chance of choosing drawer 3.
Pr(D1/S)=0(bc there are no silver coins in drawer 1)
Pr(D2/S)=0.5 (bc there is only 1 silver coin out of two coins in drawer 2)
Pr(D3/S)=1.0 (bc there are two coins in drawer 3 and both are silver)

P(S/D3)= [1*(1/3)] / [0+(.5*1/3) + (1*1/3)] =0.67% that it will come from drawer 3.

Am I close?

 

[Ray Vickson]

Homework Statement

I am just wanting to get a good starting point... not an answer to the question. Question: A desk contains three drawers. Drawer 1 has two gold coins. Drawer 3 has one gold coin and one silver coin. Drawer 3 has two silver coins. I randomly choose a drawer and then randomly choose a coin. If a silver coin is selected, what is the probability that i choose drawer 3? I am suppose to use Bayes rule to find probability. Thank you for any help

Homework Equations

Bayes Rule that I used: P(S/D3)= [P(D3/S)P(D3)] / [P(D1/S)P(S) + P(D2/S)P(S)+P(D3/S)P(S)]

The Attempt at a Solution

I defined all I could: Drawer(D), Silver(S) and Gold(G).
P(D1)=P(D2)=P(D3)= 1/3 b/c he has a one in three chance of choosing drawer 3.
Pr(D1/S)=0(bc there are no silver coins in drawer 1)
Pr(D2/S)=0.5 (bc there is only 1 silver coin out of two coins in drawer 2)
Pr(D3/S)=1.0 (bc there are two coins in drawer 3 and both are silver)

P(S/D3)= [1*(1/3)] / [0+(.5*1/3) + (1*1/3)] =0.67% that it will come from drawer 3.

Am I close?

No, you are not. P(S|D3) is obtainable directly from the data (drawer 3 has one gold and one silver coin). You were not asked to find P(S|D3); you want to know P(D3|S). Also, 0.67% = 0.0067 ≈ 2/300. Maybe you meant 67%.

RGV