Respuesta :
Answer:
49.67% probability that he makes an ace on at least 2 of the 8 serves.
Step-by-step explanation:
For each serve, there are only two possible outcomes. Either the player makes an ace, or he does not. The probability of the player making an ace in a serve is independent from other serves. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
It is believed that the probability of serving an ace in tennis on any given serve is 0.2.
This means that [tex]p = 0.2[/tex]
If a tennis player serves 8 times, what is the probability that he makes an ace on at least 2 of the 8 serves?
This is [tex]P(X \geq 2)[/tex] when [tex]n = 8[/tex].
We know that either he makes an ace on less than 2 serves, or he makes an ace in at least 2 of the serves. The sum of the probabilities of these events is 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{8,0}.(0.2)^{0}.(0.8)^{8} = 0.1678[/tex]
[tex]P(X = 1) = C_{8,1}.(0.2)^{1}.(0.8)^{7} = 0.3355[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1678 + 0.3355 = 0.5033[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5033 = 0.4967[/tex]
49.67% probability that he makes an ace on at least 2 of the 8 serves.
