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Question

From a bag containing n balls, all either white or black, all numbers of each being equally likely, a ball is drawn which turns out to be white; this is replaced, and another ball is drawn, which also turns out to be white. If this ball is replaced, prove that the chance of the next draw giving a black ball is $$\displaystyle\frac{1}{2}(n-1)(2n+1)^{-1}$$.


Solution

We have n cases to consider, for there may be $$1,2,3,...n$$ white balls; and all these cases are equally likely,
so that
$${ P }_{ 1 }={ P }_{ 2 }={ P }_{ 3 }=.....={ P }_{ n }$$
If there were r white balls, the chance of drawing two white balls in this case should be $${ \left( \cfrac { r }{ n }  \right)  }^{ 2 }$$
$$\therefore \cfrac { { Q }_{ 1 } }{ { \left( \cfrac { 1 }{ n }  \right)  }^{ 2 } } =\cfrac { { Q }_{ 2 } }{ { \left( \cfrac { 2 }{ n }  \right)  }^{ 2 } } =....\cfrac { { Q }_{ r } }{ { \left( \cfrac { r }{ n }  \right)  }^{ 2 } } =.....=\cfrac { 1 }{ \sum { \cfrac { { r }^{ 2 } }{ { n }^{ 2 } }  }  } $$
Thus $$\cfrac { { Q }_{ r } }{ { \left( \cfrac { r }{ n }  \right)  }^{ 2 } } =\cfrac { 6n }{ \left( n+1 \right) \left( 2n+1 \right)  } ,{ Q }_{ r }=\cfrac { 6{ r }^{ 2 } }{ n(n+1)(2n+1) } $$
And the chance of another drawung giving a black ball
$$\sum _{ r=1 }^{ r=n }{ \cfrac { n-r }{ n }  } .\cfrac { 6{ r }^{ 2 } }{ n(n+1)(2n+1) } =\cfrac { 6\sum { { r }^{ 2 } }  }{ n(n+1)(2n+1) } -\cfrac { 6\sum { { r }^{ 3 } }  }{ { n }^{ 2 }\left( n+1 \right) \left( 2n+1 \right)  } $$
$$=1-\cfrac { 6{ n }^{ 2 }{ (n+1) }^{ 2 } }{ 4{ n }^{ 2 }(n+1)(2n+1) } =1-\cfrac { 3(n+1) }{ 2(2n+1) } =\cfrac { 1 }{ 2 } \left( n-1 \right) { (2n+1) }^{ -1 }$$

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