It has been determined that 5% of drivers checked at a road stop show traces of alcohol and 10% of drivers checked do not wear seat belts. In addition, it has been observed that the two infractions are independent from one another. If an officer stops five drivers at random:
- Calculate the probability that exactly three of the drivers have committed any one of the two offenses.
- Calculate the probability that at least one of the drivers checked has committed at least one of the two offenses
My doubts: If we only had one of the events instead of two, this would be a standard binomial distribution problem. How does having two observations change this problem?
Solution:
- It's tedious but:
P(3 infarctions)= P(3 drunks)+P(3 with no belts)+P(2 drunks, 1 with no belt)+ P(1 drunk, 2 with no belt)
The problem I have is, how to actually compute
P(3 are drunk)
Is it 5C3 * (0.05)3* (0.9)5*(0.95)2 ?
Or something else ?
I break this down into one driver having 4 cases:
1-They're drunk, They don't have seat belts
2-They're not drunk, They don't have seat belts
3-They're drunk, they have seat belts on
4-They're not drunk, they have seat belts on
Is the above interpretation correct ?
Probabilities then can be broken down into two components. Which is what I did for the case of
P(3 are drunk)
= 5C3 *(This driver has one infarction)*3 *(This driver has no infarctions)
There is also debate for the last one. It's solution is simple, find the probability that none of the drivers have committed any infarctions, and subtract it with 1.
Is it :
P(no infarctions)= (0.9)5*(0.95)5
One driver has this : (0.9)*(0.95) probability of not having committed any of the infarctions, there are 5 drivers so their probabilities will be multiplied.
Or will there be a variation of it?
Best Answer
We can reduce both problems to binomial distribution problems in one variable. Looking at a single driver, the probability of them being either drunk or not wearing a seatbelt is $$(0.05)(1-0.1)+(1-0.05)(0.1)=0.14$$ And the probability of them being drunk or not wearing a seatbelt or both is $$1-(1-0.05)(1-0.1)=0.145$$ Let $X$ and $Y$ be the random variables representing the number of drivers who satisfy the first and second of these conditions respectively. They are also binomial random variables with $n=5$, and $p=0.14$ and $p=0.145$ respectively.
With this simplification you should be able to solve both problems yourself.