Adela Brown

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

Beginner2021-12-13Added 26 answers

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$n=10\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}p=0.44$ .

Probability mass function of Binomial variable X is:

$P(X=x){=}^{n}{C}_{x}{p}^{x}{(1-p)}^{n-x}$

where,$n=0,1,2\dots 10$

Step 2

Required probability:

$P(X\le 1)=P(X=0)+P(X=1)$

$=}^{10}{C}_{0}{\left(0.44\right)}^{0}{(1-0.44)}^{10-0}{+}^{10}{C}_{1}{\left(0.44\right)}^{1}{(1-0.44)}^{10-1$

$=\frac{10!}{0!(10-0)!}\times 1\times 0.003033055+\frac{10!}{1!(10-1)!}\times 0.44\times 0.00541617$

$=1\times 0.003033055+10\times 0.0023831148$

$=0.003033055+0.023831148$

$=0.026864203$

$\approx 0.0269$

Thus, the probability that the day will end with at most one speeding tickets is 0.0269

Given Information:

Garam Doe’s daily chores require making 10 round trips by car between two towns. i.e., number of trials

There is a 44% chance of getting a speeding fine on any round trip. i.e.,

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:

where,

Step 2

Required probability:

Thus, the probability that the day will end with at most one speeding tickets is 0.0269

amarantha41

Beginner2021-12-14Added 38 answers

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$\left(n\right)=10\text{}days$

The process will follow a binomial distribution with a probability distribution function as follows:

$P(X=x){=}^{n}{C}_{x}\cdot \left({p}^{x}\right)\cdot {(1-p)}^{n-x}$

$P(X=0)=10{C}_{0}\cdot \left({0.44}^{0}\right)\cdot {(1-0.44)}^{10-0}=0.0030$

$P(X=1)=10{C}_{1}\cdot \left({0.44}^{1}\right)\cdot {(1-0.44)}^{10-1}=0.0238$

What is the probability that the day will end with at most one speeding tickets (one or less) (i.e. 0 and 1)

$=P(X=0)+P(X=1)$

$=0.0030+0.0238$

$=0.0268$

Every day there is a 44% chance of getting a speeding fine on any round trip.

Sample size

The process will follow a binomial distribution with a probability distribution function as follows:

What is the probability that the day will end with at most one speeding tickets (one or less) (i.e. 0 and 1)

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