2021-12-12

Garam Doe’s daily chores require making 10 round trips by car between two towns. Once through with all ten trips, Mr. Doe can take the rest of the day off, a good enough motivation to drive above the speed limit. Experience shows that there is a 44% chance of getting a speeding fine on any round trip. What is the probability that the day will end with at most one speeding tickets (one or less)

raefx88y

Step 1
Given Information:
Garam Doe’s daily chores require making 10 round trips by car between two towns. i.e., number of trials $n=10$
There is a 44% chance of getting a speeding fine on any round trip. i.e., $p=0.44$
We are asked to find the probability that the day will end with at most one speeding tickets:
Let X denote the number of speeding tickets and X follows Binomial distribution with parameters .
Probability mass function of Binomial variable X is:
$P\left(X=x\right){=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{n-x}$
where, $n=0,1,2\dots 10$
Step 2
Required probability:
$P\left(X\le 1\right)=P\left(X=0\right)+P\left(X=1\right)$
${=}^{10}{C}_{0}{\left(0.44\right)}^{0}{\left(1-0.44\right)}^{10-0}{+}^{10}{C}_{1}{\left(0.44\right)}^{1}{\left(1-0.44\right)}^{10-1}$
$=\frac{10!}{0!\left(10-0\right)!}×1×0.003033055+\frac{10!}{1!\left(10-1\right)!}×0.44×0.00541617$
$=1×0.003033055+10×0.0023831148$
$=0.003033055+0.023831148$
$=0.026864203$
$\approx 0.0269$
Thus, the probability that the day will end with at most one speeding tickets is 0.0269

amarantha41

From the provided information,
Every day there is a 44% chance of getting a speeding fine on any round trip. $p=0.44$.
Sample size
The process will follow a binomial distribution with a probability distribution function as follows:
$P\left(X=x\right){=}^{n}{C}_{x}\cdot \left({p}^{x}\right)\cdot {\left(1-p\right)}^{n-x}$
$P\left(X=0\right)=10{C}_{0}\cdot \left({0.44}^{0}\right)\cdot {\left(1-0.44\right)}^{10-0}=0.0030$
$P\left(X=1\right)=10{C}_{1}\cdot \left({0.44}^{1}\right)\cdot {\left(1-0.44\right)}^{10-1}=0.0238$
What is the probability that the day will end with at most one speeding tickets (one or less) (i.e. 0 and 1)
$=P\left(X=0\right)+P\left(X=1\right)$
$=0.0030+0.0238$
$=0.0268$

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