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December 18, 2014

December 18, 2014

Posted by **marie** on Thursday, September 25, 2008 at 4:50pm.

- statistics -
**David Q**, Thursday, September 25, 2008 at 5:29pmI don't have a binomial probabilities table, so I've done the math from scratch: you should be able to verify the probabilities against your tables as you go:

The probability of a driver being drunk is 0.2, so the probability of a driver NOT being drunk is 1 - 0.2 = 0.8. So the probability of all 12 drivers not being drunk is (0.8)^12 = 0.0687.

The probability of exactly one being drunk is 12*(0.2)*((0.8)^11) (because there are 12 different ways in which one driver out of the 12 can be drunk).

The probability of exactly two being drunk is (12*11/2)*((0.2)^2)*((0.8)^10) (because there are 12*11/2 ways in which this can happen).

The probability of exactly three being drunk is (12*11*10/(3*2))*((0.2)^3)*((0.8)^9) (because there are 12*11*10/(3*2) ways in which this can happen).

The probability of exactly four being drunk is (12*11*10*9/(4*3*2))*((0.2)^4)*((0.8)^8) (because there are 12*11*10/(3*2) ways in which this can happen).

In general, the probability of there being K drunk drivers out of a total of N is:

Fact(N)/(Fact(K)*Fact(N-K))

*((0.2)^K)*((0.8)^(N-K))

where Fact(K) means K*(K-1)*(K-2)*...*1, so you should be able to check that the above formula gives the same answers for 0, 1, 2, 3 and 4 drunks as written above.

You can work out the probability of there being exactly 5, 6, ..., 12 drunk drivers in your sample by working out all the remaining terms in this sequence (and you ought really to do that, just to check that the total probability of all the possible outcomes equals 1.0), but you now have all the information you need to answer the question.

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