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When the Bucks play Chiefs at football, the probability that the Chiefs, on present form, will win is 0.56. In a competition, these teams are to play two more pgames. If Swallows beats Bucks in at least4one of these games, they will win the competition, otherwise Bucks will win the trophy. NB: Round off to 2 decimal places. a. The probability that Swallows will win the trophy is [a] probability that Rucks will win the trophy is

User Adeina
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16 votes

Answer:

The probability that Swallows will win the trophy is 0.8064

The probability that Rucks will win the trophy is 0.1936

Explanation:

For each game, there are only two possible outcomes. Either the Swallows win, or they do not. The probability of them winning a game is independent of any other game, which means that the binomial probability distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.


P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)

In which
C_(n,x) is the number of different combinations of x objects from a set of n elements, given by the following formula.


C_(n,x) = (n!)/(x!(n-x)!)

And p is the probability of X happening.

Probability the Swallows wins is 0.56

This means that
p = 0.56

2 games:

This means that
n = 2

The probability that Swallows will win the trophy is

Probability they win at least one game, so:


P(X \geq 1) = 1 - P(X = 0)

In which


P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)


P(X = 0) = C_(2,0).(0.56)^(0).(0.44)^(2) = 0.1936

Then


P(X \geq 1) = 1 - 0.1936 = 0.8064

0.8064 = 80.64% probability the Swallows win the trophy and 0.1936 probability that the Rucks win the trophy.

User Ndasusers
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