  Question

$$A$$ and $$B$$ play a game in which $$A'$$s chance of winning is $$1/5$$. In a series of $$6$$ games, the probability that $$A$$ will win at least three games is

A
63C(45)3(15)3+64C(45)2(15)4  B
63C(45)3(15)3  C
63C(45)3(15)3+64C(45)2(15)4+65C(45)1(15)5+(15)6  D
63C(45)3(15)3+64C(45)2(15)4+65C(45)1(15)5  Solution

The correct option is C $$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }+_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 5 }+{ \left( \cfrac { 1 }{ 5 } \right) }^{ 6 }$$Probability of $$A$$ winning the game is $$\displaystyle P(A)=(\frac{1}{5})$$,Probability of $$B$$ winning the game is $$\displaystyle P(B)=(\frac{4}{5})$$Any three games out of total $$6$$ games can be chosen in $$^{ 6 }C_{ 3 }$$ ways.Probability of $$A$$ winning any three particular games and losing the remaining is $$\displaystyle { \left( \frac { 1 }{ 5 } \right) }^{ 3 }{ \left( \frac { 4 }{ 5 } \right) }^{ 3 }$$Hence probability that $$A$$ will win exactly any $$3$$ of the $$6$$ games, using multiplication theorem is $$P(A=3) =\displaystyle {}^{ 6 }C_{ 3 }{ \left( \frac { 1 }{ 5 } \right) }^{ 3 }{ \left( \frac { 4 }{ 5 } \right) }^{ 3 }$$Similarly probability that $$A$$ will win exactly $$4, 5, 6$$ games are $$\displaystyle ^{ 6 }C_{ 4 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }{ \left( \frac {4 }{ 5 } \right) }^{ 2 },\; ^{ 6 }C_{ 5 }{ \left( \frac { 1 }{ 5 } \right) }^{ 5 }{ \left( \frac { 4 }{ 5 } \right) }^{ 1 },\; ^{ 6 }C_{6 }{ \left( \frac { 1 }{ 5 } \right) }^{ 6 }$$ respectively.Hence the probability of $$A$$ winning at least $$3$$ games $$\displaystyle = P(A=3)+P(A=4)+P(A=5)+P(A=6) \\ \displaystyle ={}^{6 }C_{ 3 }{ \left( \frac { 1 }{ 5 } \right) }^{ 3 }{ \left( \frac { 4 }{ 5 } \right) }^{ 3 }+{}^{ 6 }C_{ 4 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 }+{}^{ 6 }C_{ 5 }{ \left( \frac { 1 }{ 5 } \right) }^{ 5 }{ \left( \frac { 4 }{ 5 } \right) }^{ 1 }+{}^{ 6 }C_{ 6 }{ \left( \frac { 1 }{ 5 } \right) }^{ 6 }$$Maths

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