$A$ 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

_{3}^{6} C {(\frac{4}{5})}^{3} {(\frac{1}{5})}^{3} +_{4}^{6} C {(\frac{4}{5})}^{2} {(\frac{1}{5})}^{4}

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_{3}^{6} C {(\frac{4}{5})}^{3} {(\frac{1}{5})}^{3}

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_{3}^{6} C {(\frac{4}{5})}^{3} {(\frac{1}{5})}^{3} +_{4}^{6} C {(\frac{4}{5})}^{2} {(\frac{1}{5})}^{4} +_{5}^{6} C {(\frac{4}{5})}^{1} {(\frac{1}{5})}^{5} + {(\frac{1}{5})}^{6}

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_{3}^{6} C {(\frac{4}{5})}^{3} {(\frac{1}{5})}^{3} +_{4}^{6} C {(\frac{4}{5})}^{2} {(\frac{1}{5})}^{4} +_{5}^{6} C {(\frac{4}{5})}^{1} {(\frac{1}{5})}^{5}

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Solution

The correct option is C $_{3}^{6} C {(\frac{4}{5})}^{3} {(\frac{1}{5})}^{3} +_{4}^{6} C {(\frac{4}{5})}^{2} {(\frac{1}{5})}^{4} +_{5}^{6} C {(\frac{4}{5})}^{1} {(\frac{1}{5})}^{5} + {(\frac{1}{5})}^{6}$
Probability of $A$ winning the game is $P (A) = (\frac{1}{5})$ ,
Probability of $B$ winning the game is $P (B) = (\frac{4}{5})$
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
${(\frac{1}{5})}^{3} {(\frac{4}{5})}^{3}$
Hence probability that $A$ will win exactly any $3$ of the $6$ games, using multiplication theorem is
$P (A = 3) =^{6} C_{3} {(\frac{1}{5})}^{3} {(\frac{4}{5})}^{3}$
Similarly probability that $A$ will win exactly $4, 5, 6$ games are $^{6} C_{4} {(\frac{1}{5})}^{4} {(\frac{4}{5})}^{2},^{6} C_{5} {(\frac{1}{5})}^{5} {(\frac{4}{5})}^{1},^{6} C_{6} {(\frac{1}{5})}^{6}$ respectively.
Hence the probability of $A$ winning at least $3$ games $= P (A = 3) + P (A = 4) + P (A = 5) + P (A = 6) =^{6} C_{3} {(\frac{1}{5})}^{3} {(\frac{4}{5})}^{3} +^{6} C_{4} {(\frac{1}{5})}^{4} {(\frac{4}{5})}^{2} +^{6} C_{5} {(\frac{1}{5})}^{5} {(\frac{4}{5})}^{1} +^{6} C_{6} {(\frac{1}{5})}^{6}$

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