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A cabinet has three drawers. In the first drawer there are 3 gold balls. In the second drawer there are 3 silver balls. In the third drawer there are 3 silver balls and 2 gold balls An experiment consists of choosing a drawer, and then choosing a ball inside it. Each drawer is equally likely to be chosen, and within each drawer, each ball is equally likely to be chosen. For example, P(A gold ball is chosen | Drawer 3 is chosen) 2/5 Given that a gold ball is chosen, what is the probability that the drawer with 3 gold balls was selected? Now what if all the balls are numbered from 1-3 based on which drawer they were in, and then taken from the drawers and placed into a large pile of 11 balls? Each ball in this pile is now equally likely to be chosen, unlike in the previous part. Given that a gold ball is chosen, what is the probability that the gold ball came from the drawer with 3 gold balls?

Respuesta :

Solution :

Let [tex]$D_1,D_2,D_3$[/tex] be the events that the first, second and the third drawers are selected respectively.

Therefore, [tex]$P(D_1)=P(D_2)=P(D_3)=\frac{1}{3}$[/tex]

Now the G shows that the events that the gold ball is selected, so

[tex]$P(G|D_1)=1,P(G|D_2)=0,P(G|D_3)=\frac{2}{5},$[/tex]

By probability, he gold ball is selected is :

[tex]$P(G)=P(G|D_1)P(D_1)+P(G|D_2)P(D_2)+P(G|D_3)P(D_3)$[/tex]

        [tex]$=1.\frac{1}{3}+0.\frac{1}{3}+\frac{2}{5}.\frac{1}{3}$[/tex]

        [tex]$=\frac{7}{15}$[/tex]

Now the required probability is :

[tex]$P(D_1|G)=\frac{P(G|D_1)P(D_1)}{P(G)}$[/tex]

               [tex]$=\frac{1/3}{7/15}$[/tex]

                [tex]$=\frac{5}{7}$[/tex]

                  = 0.71

Now out of 11 balls, 3+2 = 5 balls are gold balls.

Therefore, [tex]$P(G)=\frac{5}{11}$[/tex]

The probability that a gold ball and the first drawer is selected is :

[tex]$P(G \text{ and first drawer})= \frac{3}{11}$[/tex]

Now the required probability is :

[tex]$P(\text{first drawer}|G)=\frac{P(\text{ G and first drawer })}{P(G)}$[/tex]

                           [tex]$=\frac{3/11}{5/11}$[/tex]

                           [tex]$=\frac{3}{5}$[/tex]

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