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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
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B
63C(45)3(15)3
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C
63C(45)3(15)3+64C(45)2(15)4+65C(45)1(15)5+(15)6
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D
63C(45)3(15)3+64C(45)2(15)4+65C(45)1(15)5
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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 $$ \displaystyle = P(A=3)+P(A=4)+P(A=5)+P(A=6) \\
 \displaystyle ={}^{6 }C_{ 3 }{ \left( \frac { 1 }{ 5 }  \right)  }^{ 3 }{ \left( \frac { 4
}{ 5 }  \right)  }^{ 3 }+{}^{ 6 }C_{ 4 }{ \left( \frac { 1 }{ 5 } 
\right)  }^{ 4 }{ \left( \frac { 4 }{ 5 }  \right)  }^{ 2 }+{}^{ 6 }C_{ 5
}{ \left( \frac { 1 }{ 5 }  \right)  }^{ 5 }{ \left( \frac { 4 }{ 5 } 
\right)  }^{ 1 }+{}^{ 6 }C_{ 6 }{ \left( \frac { 1 }{ 5 }  \right)  }^{ 6 }$$

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