A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If there b

Question

A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If there b

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

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`