Respuesta :
Answer:
A) 0.1612
B) 0.8031
C) 0.1969
Step-by-step explanation:
For each toss of the die, there are only two possible outcomes. Either it is a 3, or it is not. The probability of getting a 3 on each toss is independent from other tosses. So we use the binomial probability distribution to solve this problem.
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.
In this problem we have that:
Five tosses, so [tex]n = 5[/tex]
The die has 6 values, from 1 to 6. The die is fair, so each outcome is equally as likely. The probability of a 3 appearing in a single throw is [tex]p = \frac{1}{6} = 0.167[/tex]
(a) twice
This is [tex]P(X = 2)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{5,2}.(0.167)^{2}.(0.833)^{3} = 0.1612[/tex]
(b) at most once
[tex]P(X \leq 1) = 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_{5,0}.(0.167)^{0}.(0.833)^{5} = 0.4011[/tex]
[tex]P(X = 1) = C_{5,1}.(0.167)^{1}.(0.833)^{1} = 0.4020[/tex]
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.4011 + 0.4020 = 0.8031[/tex]
(c) at least two times.
Either a 3 appears at most once, or it does at least two times. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 1) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.8031 = 0.1969[/tex]
