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Munni Ajay Tata
Asked: 2022-11-08T11:58:40+05:30 In:

A bag contains some white and some black balls, all combinations ofballs being equally likely. The total number of balls in the bag is 10. Ifthere ball are drawn at random without replacement and all of them are foundto be black, the probability that eh bag contains 1 white and 9 black ballsis`14//55`b. `12//55`c. `2//11`d. `8//55`
A. `14//55`
B. `15//55`
C. `2//11`
D. `8//55`

A bag contains some white and some black balls, all combinations ofballs being equally likely. The total number of balls in the bag is 10. Ifthere ball are drawn at random without replacement and all of them are foundto be black, the probability that eh bag contains 1 white and 9 black ballsis`14//55`b. `12//55`c. `2//11`d. `8//55`
A. `14//55`
B. `15//55`
C. `2//11`
D. `8//55`
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  1. Correct Answer – A
    Let `E_(i)` denote the event that the bag contains I blck and `(10-i)` white balls `(i=0,1,2,…,10).` Let A denote the event that the three balls drawn at random from the bag are black. We have,
    `P(E_(i))=1/11(i=0,1,2…,10)`
    `P(A//E_(i))=0″for” i=0,1,2and P(A//E_(i))=(“”^(i)C_(3))/(“”^(10)C_(3))”for”ige3`
    `impliesP(A)=1/11xx(1)/(“”^(10)C_(3))[“”^(3)C_(3)+””^(4)C_(3)+…+””^(10)C_(3)]`
    But `””^(3)C_(3)+””^(4)C_(3)+””^(5)C_(3)+…+””^(10)C_(3)=””^(4)C_(4)+””^(4)C_(3)+””^(5)C_(3)+…+””^(10)C_(3)`
    `” “=””^(5)C_(4)+””^(5)C_(3)+””^(6)C_(3)+…+””^(10)C_(3)=””^(11)C_(4)`
    `impliesP(A)=1/11xx(1)/(“”^(10)C_(3))xx””^(11)C_(4)=((11xx10xx9xx8)/(4!))/(11xx(10xx9xx8)/(3!))=1/4`
    `thereforeP(E_(9)//A)=(P(E_(9))P(A//E_(9)))/(P(A))=((1)/(11)xx(“”^(9)C_(3))/(“”^(10)C_(3)))/(1/4)=14/55`

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