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Box-I-has-3-red-and-5-white-balls-while-Box-II-contains-4-red-and-2-white-balls-A-ball-is-chosen-at-random-from-the-first-box-and-placed-in-the-second-box-without-observing-its-colour-Then-a-ball-

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Question Number 3824 by Yozzii last updated on 21/Dec/15
Box I has 3 red and 5 white balls, while Box II contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its colour. Then a ball is drawn from the second box. Find the probability that it is white.
$${Box}\:{I}\:{has}\:\mathrm{3}\:{red}\:{and}\:\mathrm{5}\:{white}\:{balls}, \\ $$$${while}\:{Box}\:{II}\:{contains}\:\mathrm{4}\:{red}\:{and}\:\mathrm{2}\: \\ $$$${white}\:{balls}.\:{A}\:{ball}\:{is}\:{chosen}\:{at}\:{random} \\ $$$${from}\:{the}\:{first}\:{box}\:{and}\:{placed}\:{in}\:{the} \\ $$$${second}\:{box}\:{without}\:{observing}\:{its}\:{colour}. \\ $$$${Then}\:{a}\:{ball}\:{is}\:{drawn}\:{from}\:{the}\:{second} \\ $$$${box}.\:{Find}\:{the}\:{probability}\:{that}\:{it}\:{is}\:{white}. \\ $$
Commented by Rasheed Soomro last updated on 21/Dec/15
Act−I :Drawing a ball from Box I Total balls=8,out of which are 5 white balls. probability of being white ball is (5/8) Act−II :4 red 2 white one white^((5/8)) /red^((3/8)) One added ball with respect to probability is (5/8) parts white and (3/8) parts red Total balls are 7 Out of which 2+(5/8)=((21)/8) white balls probability being white=(((21)/8)/7)=((21)/(8×7))=(3/8)
$${Act}−{I}\:\::{Drawing}\:{a}\:{ball}\:{from}\:{Box}\:{I} \\ $$$${Total}\:{balls}=\mathrm{8},{out}\:{of}\:{which}\:{are}\:\mathrm{5}\:{white} \\ $$$${balls}. \\ $$$${probability}\:{of}\:{being}\:{white}\:{ball}\:{is}\:\frac{\mathrm{5}}{\mathrm{8}} \\ $$$${Act}−{II}\:\::\mathrm{4}\:{red}\:\mathrm{2}\:{white}\:{one}\:\overset{\frac{\mathrm{5}}{\mathrm{8}}} {{white}}/\overset{\frac{\mathrm{3}}{\mathrm{8}}} {{red}} \\ $$$${One}\:{added}\:{ball}\:{with}\:{respect}\:{to}\:{probability} \\ $$$${is}\:\frac{\mathrm{5}}{\mathrm{8}}\:{parts}\:{white}\:{and}\:\frac{\mathrm{3}}{\mathrm{8}}\:{parts}\:{red} \\ $$$${Total}\:{balls}\:{are}\:\mathrm{7} \\ $$$${Out}\:{of}\:{which}\:\mathrm{2}+\frac{\mathrm{5}}{\mathrm{8}}=\frac{\mathrm{21}}{\mathrm{8}}\:{white}\:{balls} \\ $$$${probability}\:{being}\:{white}=\frac{\frac{\mathrm{21}}{\mathrm{8}}}{\mathrm{7}}=\frac{\mathrm{21}}{\mathrm{8}×\mathrm{7}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by prakash jain last updated on 21/Dec/15
E_1 =White ball drawn from Box1 E_2 =Red ball drawn from Box1 A=getting white ball from Box2 P(A)=P(A∣E_1 )P(E_1 )+P(A∣E_2 )P(E_2 ) P(E_1 )=5/8 P(E_2 )=3/8 P(A∣E_1 )=3/7 P(A∣E_2 )=2/7 =(3/7)×(5/8)+(2/7)×(3/8)=((15+6)/(56))=((21)/(56))=(3/8)
$${E}_{\mathrm{1}} =\mathrm{White}\:\mathrm{ball}\:\mathrm{drawn}\:\mathrm{from}\:\mathrm{Box1} \\ $$$${E}_{\mathrm{2}} =\mathrm{Red}\:\mathrm{ball}\:\mathrm{drawn}\:\mathrm{from}\:\mathrm{Box1} \\ $$$${A}=\mathrm{getting}\:\mathrm{white}\:\mathrm{ball}\:\mathrm{from}\:\mathrm{Box2} \\ $$$$\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{A}\mid\mathrm{E}_{\mathrm{1}} \right)\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \right)+\mathrm{P}\left(\mathrm{A}\mid\mathrm{E}_{\mathrm{2}} \right)\mathrm{P}\left(\mathrm{E}_{\mathrm{2}} \right) \\ $$$$\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \right)=\mathrm{5}/\mathrm{8} \\ $$$$\mathrm{P}\left(\mathrm{E}_{\mathrm{2}} \right)=\mathrm{3}/\mathrm{8} \\ $$$$\mathrm{P}\left(\mathrm{A}\mid\mathrm{E}_{\mathrm{1}} \right)=\mathrm{3}/\mathrm{7} \\ $$$$\mathrm{P}\left(\mathrm{A}\mid\mathrm{E}_{\mathrm{2}} \right)=\mathrm{2}/\mathrm{7} \\ $$$$=\frac{\mathrm{3}}{\mathrm{7}}×\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{2}}{\mathrm{7}}×\frac{\mathrm{3}}{\mathrm{8}}=\frac{\mathrm{15}+\mathrm{6}}{\mathrm{56}}=\frac{\mathrm{21}}{\mathrm{56}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by Yozzii last updated on 21/Dec/15
Thanks
$${Thanks} \\ $$

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