Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 9662 by Joel575 last updated on 23/Dec/16

I have 2 buckets. Each bucket contains green and blue balls  The first bucket contains 3 green balls and 7 blue balls.  Second bucket contains 7 green balls and 8 blue balls.  I want to take those balls with coin toss.  If head, I will take 1 ball from each bucket.  But if tail, I will take 2 balls from each bucket.  What is the propability if all the balls that have been taken  have the same color?  (sorry for my grammar)

$$\mathrm{I}\:\mathrm{have}\:\mathrm{2}\:\mathrm{buckets}.\:\mathrm{Each}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{green}\:\mathrm{and}\:\mathrm{blue}\:\mathrm{balls} \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{3}\:\mathrm{green}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{7}\:\mathrm{blue}\:\mathrm{balls}. \\ $$$$\mathrm{Second}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{7}\:\mathrm{green}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{8}\:\mathrm{blue}\:\mathrm{balls}. \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{take}\:\mathrm{those}\:\mathrm{balls}\:\mathrm{with}\:\mathrm{coin}\:\mathrm{toss}. \\ $$$$\mathrm{If}\:\mathrm{head},\:\mathrm{I}\:\mathrm{will}\:\mathrm{take}\:\mathrm{1}\:\mathrm{ball}\:\mathrm{from}\:\mathrm{each}\:\mathrm{bucket}. \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{tail},\:\mathrm{I}\:\mathrm{will}\:\mathrm{take}\:\mathrm{2}\:\mathrm{balls}\:\mathrm{from}\:\mathrm{each}\:\mathrm{bucket}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{propability}\:\mathrm{if}\:\mathrm{all}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{that}\:\mathrm{have}\:\mathrm{been}\:\mathrm{taken} \\ $$$$\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{color}? \\ $$$$\left({sorry}\:{for}\:{my}\:{grammar}\right) \\ $$

Answered by sandy_suhendra last updated on 23/Dec/16

we will get  (i) head,green,green=(1/2)×(3/(10))×(7/(15))=((21)/(300))  (ii) head,blue,blue=(1/2)×(7/(10))×(8/(15))=((56)/(300))  (iii) tail,(g,g),(g,g)=(1/2)×((3C2)/(10C2))×((7C2)/(15C2))=(2/(300))  (iv) tail,(b,b),(b,b)=(1/2)×((7C2)/(10C2))×((8C2)/(15C2))=((14)/(225))  so the probability=((21)/(300))+((56)/(300))+(2/(300))+((14)/(225))=((293)/(900))

$$\mathrm{we}\:\mathrm{will}\:\mathrm{get} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{head},\mathrm{green},\mathrm{green}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{10}}×\frac{\mathrm{7}}{\mathrm{15}}=\frac{\mathrm{21}}{\mathrm{300}} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{head},\mathrm{blue},\mathrm{blue}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{7}}{\mathrm{10}}×\frac{\mathrm{8}}{\mathrm{15}}=\frac{\mathrm{56}}{\mathrm{300}} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{tail},\left(\mathrm{g},\mathrm{g}\right),\left(\mathrm{g},\mathrm{g}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3C2}}{\mathrm{10C2}}×\frac{\mathrm{7C2}}{\mathrm{15C2}}=\frac{\mathrm{2}}{\mathrm{300}} \\ $$$$\left(\mathrm{iv}\right)\:\mathrm{tail},\left(\mathrm{b},\mathrm{b}\right),\left(\mathrm{b},\mathrm{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{7C2}}{\mathrm{10C2}}×\frac{\mathrm{8C2}}{\mathrm{15C2}}=\frac{\mathrm{14}}{\mathrm{225}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{probability}=\frac{\mathrm{21}}{\mathrm{300}}+\frac{\mathrm{56}}{\mathrm{300}}+\frac{\mathrm{2}}{\mathrm{300}}+\frac{\mathrm{14}}{\mathrm{225}}=\frac{\mathrm{293}}{\mathrm{900}}\:\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com