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Question

A and B play a game in which As 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 63C(45)3(15)3+64C(45)2(15)4+65C(45)1(15)5+(15)6
Probability of A winning the game is P(A)=(15),
Probability of B winning the game is P(B)=(45)
Any three games out of total 6 games can be chosen in 6C3 ways.
Probability of A winning any three particular games and losing the remaining is
(15)3(45)3
Hence probability that A will win exactly any 3 of the 6 games, using multiplication theorem is
P(A=3)=6C3(15)3(45)3
Similarly probability that A will win exactly 4,5,6 games are 6C4(15)4(45)2,6C5(15)5(45)1,6C6(15)6 respectively.
Hence the probability of A winning at least 3 games =P(A=3)+P(A=4)+P(A=5)+P(A=6)=6C3(15)3(45)3+6C4(15)4(45)2+6C5(15)5(45)1+6C6(15)6

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