Question

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

Answer 1

Answer:

The probability that Swallows will win the trophy is 0.8064

The probability that Rucks will win the trophy is 0.1936

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.