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Q: What is the probability that the killers will win there first game if they only win one out 5 games?

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The probability of winning two games with the same probability of 0.8 can be calculated by multiplying the probability of winning the first game (0.8) by the probability of winning the second game (0.8). Therefore, the probability is 0.8 * 0.8 = 0.64, or 64%.

The probability that he will not win both games is 0.58

Snap, bridge games use probability.

Black Jack, Poker, any dice game, Probhex, however it can also be educational than other probability games.

That there is some element of the game whose outcome is random. Beyond that, it depends on the game.

Farmer Vaughn played in 2 games at first base for the Cincinnati Kelly's Killers in 1891, starting in none of them. He made 8 putouts, had no assists, and committed no errors, equivalent to 0 errors per game (estimate based on total games played in). He had one double play.

Jerry Hurley played in just one game at first base for the Cincinnati Kelly's Killers in 1891 and did not start. He made 6 putouts, had no assists, and committed no errors, equivalent to 0 errors per game (estimate based on total games played in). He had no double plays.

Jack Carney played in 99 games at first base for the Cincinnati Kelly's Killers in 1891, starting in none of them. He made 984 putouts, had 45 assists, and committed 28 errors, equivalent to .283 errors per game (estimate based on total games played in). He had 42 double plays.

King Kelly played in 5 games at first base for the Cincinnati Kelly's Killers in 1891, starting in none of them. He made 27 putouts, had no assists, and committed 4 errors, equivalent to .8 errors per game (estimate based on total games played in). He had one double play.

Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)However, the assumption of constant and independent probability of a win is rubbish.Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)However, the assumption of constant and independent probability of a win is rubbish.Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)However, the assumption of constant and independent probability of a win is rubbish.Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)However, the assumption of constant and independent probability of a win is rubbish.

Experimental probability

The Strata company. A sequal to the game Time Killers