Question

Hope Amelia Solo, the American soccer goalkeeper, World Cup champion and two-time Olympic gold medalist, allows goals at a rate of 0.3 per match. Let X~Pois(0.3). In her next match, what is the probability she allows more than one goal?

Answer 1

Answer:

3.70% probability she allows more than one goal

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

X~Pois(0.3)

This means that $\mu = 0.3$

In her next match, what is the probability she allows more than one goal?

Either she allows at most one goal, or she allows more than one goal. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 1) + P(X > 1) = 1$

We want $P(X > 1)$. So

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.3}*(0.3)^{0}}{(0)!} = 0.7408$

$P(X = 1) = \frac{e^{-0.3}*(0.3)^{1}}{(1)!} = 0.2222$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.7408 + 0.2222 = 0.9630$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9630 = 0.0370$

3.70% probability she allows more than one goal