MathMathter
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- Joined
- Nov 23, 2013
- Messages
- 21
Hi! I'm looking for help with a particular problem using the Binomial Probability Distribution:
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 5 drivers at random, calculate the probability that exactly 3 of the drivers have committed an offense (either offense or both offenses).
Calling the event that a driver shows traces of alcohol 'A' and calling the event that a driver is not wearing a seat belt 'B', I thought this could be solved in one of two ways:
Method 1:
First find P(A U B), then use the Binomial Probability Distribution to determine the probability of exactly 3 successes in 5 trials using p = P(A U B):
P(A U B) = .05 + .10 - (.05)(.10) = .145
P[3,5] = (5C3)(.145^3)(.855^2) = .022 or 2.2%
Method 2:
First use the Binomial Probability Distribution to determine the probability of exactly 3 successes in 5 trials for (1) p = .05 and for (2) p = .10, then find the union of the two results:
P[3,5] using p = .05: (5C3)(.05^3)(.95^2) = .0011
P[3,5] using p = .10: (5C3)(.10^3)(.90^2) = .0081
P(.0011 U .0081) = .0011 + .0081 - (.0011)(.0081) = .009 or .9%
I don't understand why I'm getting two different results though. Can someone tell me what I'm missing? Thanks!
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 5 drivers at random, calculate the probability that exactly 3 of the drivers have committed an offense (either offense or both offenses).
Calling the event that a driver shows traces of alcohol 'A' and calling the event that a driver is not wearing a seat belt 'B', I thought this could be solved in one of two ways:
Method 1:
First find P(A U B), then use the Binomial Probability Distribution to determine the probability of exactly 3 successes in 5 trials using p = P(A U B):
P(A U B) = .05 + .10 - (.05)(.10) = .145
P[3,5] = (5C3)(.145^3)(.855^2) = .022 or 2.2%
Method 2:
First use the Binomial Probability Distribution to determine the probability of exactly 3 successes in 5 trials for (1) p = .05 and for (2) p = .10, then find the union of the two results:
P[3,5] using p = .05: (5C3)(.05^3)(.95^2) = .0011
P[3,5] using p = .10: (5C3)(.10^3)(.90^2) = .0081
P(.0011 U .0081) = .0011 + .0081 - (.0011)(.0081) = .009 or .9%
I don't understand why I'm getting two different results though. Can someone tell me what I'm missing? Thanks!
