A cabinet has three drawers. In the first drawer, there are 9| gold balls. In th

Question

A cabinet has three drawers. In the first drawer, there are 9| gold balls. In the second drawer, there are 6| silver balls. In the third drawer, there are 5| silver balls and 4| 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 I Drawer 3 is chosen) = 4/9.| Given that a gold ball is chosen, what is the probability that the drawer with 9l 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 24| 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 9| gold balls?

Explanation / Answer

1)let probabilty to choose drawer 1=P(1), and son on

here let probababilty of gold ball =P(G)=P(1)*P(G|1)+P(2)*P(G|2)+P(3)*P(G|3)

=(1/3)*(9/9)+(1/2)*(0/6)+(1/3)*(4/9)=13/27

therefore Probabilty that 1st drawer chosen, given gold ball=P(1|G)=P(1)*P(G|1)/P(G) =(1/3)*(9/9)/(13/27)=9/13

2)there are 13 gold balll out of which 9 came from drawer 1.

hence probability of it is from drawer 1, given gold ball chosen =9/13

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