Garam Doe’s daily chores require making 10 round trips by ca

Adela Brown

Adela Brown

Answered question

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)

Answer & Explanation

raefx88y

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 and p=0.44.
Probability mass function of Binomial variable X is:
P(X=x)=nCxpx(1p)nx
where, n=0,1,210
Step 2
Required probability:
P(X1)=P(X=0)+P(X=1)
=10C0(0.44)0(10.44)100+10C1(0.44)1(10.44)101
=10!0!(100)!×1×0.003033055+10!1!(101)!×0.44×0.00541617
=1×0.003033055+10×0.0023831148
=0.003033055+0.023831148
=0.026864203
0.0269
Thus, the probability that the day will end with at most one speeding tickets is 0.0269
amarantha41

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 (n)=10 days
The process will follow a binomial distribution with a probability distribution function as follows:
P(X=x)=nCx(px)(1p)nx
P(X=0)=10C0(0.440)(10.44)100=0.0030
P(X=1)=10C1(0.441)(10.44)101=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

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