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$$\sum _{ r=1 }^{ r=n }{ \cfrac { n-r }{ n } } .\cfrac { 6{ r }^{ 2 } }{ n(n+1)(2n+1) } =\cfrac { 6\sum { { r }^{ 2 } } }{ n(n+1)(2n+1) } -\cfrac { 6\sum { { r }^{ 3 } } }{ { n }^{ 2 }\left( n+1 \right) \left( 2n+1 \right) }$$$$=1-\cfrac { 6{ n }^{ 2 }{ (n+1) }^{ 2 } }{ 4{ n }^{ 2 }(n+1)(2n+1) } =1-\cfrac { 3(n+1) }{ 2(2n+1) } =\cfrac { 1 }{ 2 } \left( n-1 \right) { (2n+1) }^{ -1 }$$Maths

Suggest Corrections
Â
0

Similar questions
View More

People also searched for
View More