Question

23 multiple choice 4 points find the indicated probability. empirical evidence suggests that 25% of florida drivers are uninsured. if four random florida drivers are involved in an accident, what is the probability that more than one of them are uninsured? 0.26 0.05 0.21 0.25 0.06

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 4$, $p=0.25$, and $1 - p = 0.75$.

Step2: Calculate the probability of 0 and 1 uninsured drivers

For $k = 0$:
$C(4,0)=\frac{4!}{0!(4 - 0)!}=1$
$P(X = 0)=C(4,0)\times(0.25)^{0}\times(0.75)^{4}=1\times1\times0.31640625 = 0.31640625$
For $k = 1$:
$C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4$
$P(X = 1)=C(4,1)\times(0.25)^{1}\times(0.75)^{3}=4\times0.25\times0.421875 = 0.421875$

Step3: Calculate the probability of more than 1 uninsured drivers

The probability of more than 1 uninsured drivers is $P(X>1)=1 - P(X\leq1)=1-(P(X = 0)+P(X = 1))$.
$P(X = 0)+P(X = 1)=0.31640625 + 0.421875=0.73828125$
$P(X>1)=1 - 0.73828125 = 0.26171875\approx0.26$

Answer:

0.26