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)

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raefx88y · Accepted Answer

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(1−p)n−x  where, n=0,1,2…10  Step 2  Required probability:  P(X≤1)=P(X=0)+P(X=1)  =10C0(0.44)0(1−0.44)10−0+10C1(0.44)1(1−0.44)10−1  =10!0!(10−0)!×1×0.003033055+10!1!(10−1)!×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 · Accepted Answer

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)⋅(1−p)n−x  P(X=0)=10C0⋅(0.440)⋅(1−0.44)10−0=0.0030  P(X=1)=10C1⋅(0.441)⋅(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

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

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