# Statistics with Bayes' theorem

I am working on a maths exercise and came across this question:

Given: A motorist which causes an accident, must submit to a blood test. Research shows that 1% of the drivers which cause crashes, driving under influence (alcohol, drugs). There is a chance of 75% that someone under influence is tested positive. However, there is a chance of 5% that someone who is sober is tested positive.

Wanted: (a) What is the probability that someone is under influence if the test was positive? (b) What is the probability that someone is sober, while the test was negative?

So i tried calculating these questions. a: ((0.75*0.01) /( 0.75 * 0.01 + 0.25 * 0.99)) = 0.029 or ((0.75*0.99) /( 0.75 * 0.99 + 0.25 * 0.01)) = 0.996

For b: ((0.05*0.01) /( 0.05 * 0.01 + 0.95 * 0.99))

Are these answers correct? My maths skills are terrible any help is appreciated,

Thanks,

Jef

Ps. My first language isn't english.

-

For a:

(is under infulence & test was positive) / ((is under influence & test was positive) + (is sober & test was positive).

It gives us: $( 0.01 \cdot 0.75) /(0.01 \cdot 0.75 + 0.99 \cdot 0.05)$. (0.99 is 100% minus 1% of influenced ones).

For b:

(is sober & test was negative) / ((is sober & test was negative) + (is under influence & test was negative))

So it is: $(0.99 \cdot 0.95) /( 0.99 \cdot 0.95 + 0.01 \cdot 0.25)$. (0.25 is 100%-75% because people under influence are tested positive with probability 75% so tested negative with probability of 100%-75%=25%).

In general: in numerator you put the probability of your specific event (like "is sober and was testes negative") and in the the denominator you put the probability of a more general event which is known/assumed as true (like "was tested positive" - you know the result of the test) - probability of this general event is a sum of probabilities of some smaller events including your specific event.

-
Thanks for your help and explanation! Appreciated very much!! – Jef Dec 22 '11 at 22:47