Respuesta :
Answer:
a) 0.324 = 32.4% probability that exactly 3 drivers text while driving if a police officer pulls over five drivers
b) 0.0276 = 2.76% probability the next driver texting while driving that the police officer pulls over is the fifth driver.
Step-by-step explanation:
For each driver pulled, there are only two possible outcomes. Either they are texting, or they are not. The probability of a driver being texting is independent of any other drivers. This means that the binomial probability distribution is used 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 estimated that 52% of drivers text while driving.
This means that [tex]p = 0.52[/tex]
A. What is the probability that exactly 3 drivers text while driving if a police officer pulls over five drivers?
This is [tex]P(X = 3)[/tex] when [tex]n = 5[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{5,3}.(0.52)^{3}.(0.48)^{2} = 0.324[/tex]
0.324 = 32.4% probability that exactly 3 drivers text while driving if a police officer pulls over five drivers.
B. What is the probability the next driver texting while driving that the police officer pulls over is the fifth driver?
None of the first four([tex]P(X = 0)[/tex] when [tex]n = 4[/tex]), and then the fifth, with 52% probability. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{4,0}.(0.52)^{0}.(0.48)^{4} = 0.0531[/tex]
0.0531*0.52 = 0.0276
0.0276 = 2.76% probability the next driver texting while driving that the police officer pulls over is the fifth driver.
