Question

Police records in the town of Saratoga show that 15 percent of the drivers stopped for speeding have invalid licenses. If 12 drivers are stopped for speeding, (a) Find the probability that none will have an invalid license. (Round your answer to 4 decimal places.) P(X = 0) (b) Find the probability that exactly one will have an invalid license. (Round your answer to 4 decimal places.) P(X = 1) (c) Find the probability that at least 2 will have invalid licenses. (Round your answer to 4 decimal places.) P(X ≥ 2)

Answer 1

Answer:

(a) P(x=0) = 0.1422

(b) P(x=1) = 0.3012

(c) P(x≥2) = 0.5566

Step-by-step explanation:

The probability that x drivers will have an invalid license follows a binomial distributions, so it is calculated as:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

Where n is equal to the 12 drivers stopped and p is the probability that a driver stopped for speeding have invalid licenses. Then, replacing values, we get:

$P(x)=\frac{12!}{x!(12-x)!}*0.15^{x}*(1-0.15)^{12-x}$

Now, the probability that none will have an invalid licenses is calculated as:

$P(x=0)=\frac{12!}{0!(12-0)!}*0.15^{0}*(1-0.15)^{12-0}\\P(x=0)=0.1422$

At the same way, the probability that exactly one will have an invalid license  is calculated as:

$P(x=1)=\frac{12!}{1!(12-1)!}*0.15^{1}*(1-0.15)^{12-1}\\P(x=1)=0.3012$

Finally, the probability that at least 2 will have invalid licenses is calculated as:

$P(x\geq 2)=1-P(x<2)\\P(x\geq 2)=1-(P(x=0)+P(x=1))\\P(x\geq 2)=1-(0.1422+0.3012)\\P(x\geq 2)=1-0.4434\\P(x\geq 2)=0.5566$

Answer 2

Answer:

a) $P(X=0) = (12C0) (0.15)^0 (1-0.15)^{12-0} =0.1422$

b) $P(X=1) = (12C1) (0.15)^1 (1-0.15)^{12-1} =0.3012$

c) $P(X \geq 2) = 1-P(X<2) = 1-P(X\leq 1)= 1-[P(X=0)+P(X=1)]$

And replacing we got:

$P(X \geq 2)= 1- [0.1422+0.3012]= 0.5566$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

Solution to the problem

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=12, p=0.15)$  

Part a

We want this probability:

$P(X=0) = (12C0) (0.15)^0 (1-0.15)^{12-0} =0.1422$

Part b

We want this probability:

$P(X=1) = (12C1) (0.15)^1 (1-0.15)^{12-1} =0.3012$

Part c

We want this probability:

$P(X \geq 2)$

And we can use the complement rule:

$P(X \geq 2) = 1-P(X<2) = 1-P(X\leq 1)= 1-[P(X=0)+P(X=1)]$

And replacing we got:

$P(X \geq 2)= 1- [0.1422+0.3012]= 0.5566$