Question

Penny will play 2 games of badminton against Monica. Penny’s chances of winning the first game is 70 % . If Penny wins the first game then the chances to win the second game is 80% but if Penny lose the first game then chances to win the second game is 50%. Find the probability that Penny winning exactly one match.

First game Penny (A) win =                       

First game Monica (A) win =    

Second game Penny (C) win if first is won=                        

Second game Penny (D) loss if first is won=                         

Second game Monica (E) win if first is won=                        

Second game Monica (F) loss if first is won=                      

Find the probability that Penny winning exactly one match.

Answer 1

First game Penny (A) win = 0.7

First game Monica (B) win = 1 - 0.7 = 0.3

Second game Penny (C) win if first is won= 0.8   

Second game Penny (D) loss if first is won= 1 - 0.8 = 0.2

Second game Monica (E) win if first is won= 0.5   

Second game Monica (F) loss if first is won= 1 - 0.5 = 0.5

Find the probability that Penny winning exactly one match.

P(Penny win exactly one game) = P(won first A)*P(lost second) + P(lost first) * P(won second)

= 0.7 * 0.2 + 0.3 * 0.5

= 0.14 + 0.15

= 0.29

Hence, 0.29 or 29% is the probability that Penny winning exactly one game.

Please comment if any doubt. Thank you.