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Three bags contain a number of red and white balls as follows Bag I : 3 red balls, Bag II : 2 red balls and 1 white balls and Bag III : 3 white balls.

Monjur Alahi

Three bags contain a number of red and white balls as follows Bag I : 3 red balls, Bag II : 2 red balls and 1 white balls and Bag III : 3 white balls. `i/6`, where i=1,2,3. What is the probability that
(i) a red ball will be selected? (ii) a white ball is selected?


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Salim Ahmed Kawsar

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Bag I : 3 red balls and 0 white ball
Bag II : 2 red balls and 1 white ball.
Bag III : 0 red ball andf 3 white balls.
Let `E_(1),E_(2) and E_(3)` be the events that bag I, bag II and bag III is selected and a ball is chosen from it.
`P(E_(1))=1/6,P(E_(2))=2/6and P(E_(3))=3/6`
(i) Let E be the event that a red ball is selected. Then, probablity that red ball will be selected
`P(E)=P(E_(1)cdotP(E//E_(1))+P(E_(2))cdotP(E//E_(2))+P(E_(3))cdotP(E//E_(3))`
`=1/6cdot1/3+2/6cdot2/3+3/6cdot0`
`=1/6+2/9+0`
`=(3+4)/18=7/18`
(ii) Let fr be event that a white ball is selected.
`thereforeP(F)=P(E_(1))cdotP(F//E_(1))+P(E_(2))cdotP(F//E_(2))+P(E_(3))cdotP(F//E_(3))`
`=1/6cdot0+2/6cdot1/3+3/6cdot1=1/9+3/6=11/18`
Note `P(F)=1-P(E)=1-7/18=11/18` [since, we know that P(E)+P(F)=1]

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