Answer :
Answer:
5/9
Step-by-step explanation:
Number of players = 3
number of times game is repeated = 4
P( any person wins a game ) = 1/3
P ( any person does not win a game ) = 1 - 1/3 = 2/3
P ( any person wins no game in 4 attempts ) = ( 2/3 )^4 = 16/81
Note : each player has equal chance of winning
Find the probability that there is at least one person who wins no games
lets represent the probability of each player not wining a game with alphabet A
A1 = player 1 wins no game
A2 = player 2 wins no game
A3 = players 3 wins no game
Applying the inclusion-exclusion formula
P( A1 ∪ A2 U A3 ) = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1 ∩ A3 ) + P( A1 ∩ A2 ∩ A3 )
where
P( A1 ∩ A2 ) = P( A1 wins all games )
P ( A1 wins all games in 4 attempts ) = ( 1/3 )^4 = 1/81
P( A1 ∩ A2 ∩ A3 ) = P ( no players wins any game in 4 attempts ) = 0
Hence
P( A1 ∪ A2 U A3 ) = 16/81 + 16/81 + 16/81 - 1/81 - 1/81 - 1/81 - 0 = 5/9
The probability should be[tex]5\div 9[/tex]
Calculation:
Number of players = 3
number of times game is repeated = 4
P( any person wins a game ) = [tex]1\div 3[/tex]
P ( any person does not win a game ) = [tex]1 - 1\div 3 = 2\div 3[/tex]
P ( any person wins no game in 4 attempts ) = [tex]( 2\div 3 )^4 = 16\div 81[/tex]
Now
A1 = player 1 wins no game
A2 = player 2 wins no game
A3 = players 3 wins no game
So,
P( A1 ∪ A2 U A3 ) = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1 ∩ A3 ) + P( A1 ∩ A2 ∩ A3 )
[tex]= 16\div 81 + 16 \div81 + 16\div 81 - 1 \div 81 - 1 \div 81 - 1 \div81 - 0 \\\\= 5 \div9[/tex]
Learn more about the probability here: https://brainly.com/question/795909?referrer=searchResults