Prob Question

Hello, stuck on this one.

Everyday you travel through 3 traffic lights.
The probability of stopping at first traffic light is 0.6. If you stop at one set the prob you have to stop at the next is 0.9. If you dont have to stop at one set then the prob you dont have to stop at next is 0.7.

Find prob that you stop at exactly one set of lights.

I thought it was (0.6*0.1*0.7) + (0.4*0.3*0.9) + (0.4*0.7*0.3), which is wrong. Where is the mistake? Thanks

Quote:

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Originally Posted by classicstrings

Hello, stuck on this one.

Everyday you travel through 3 traffic lights.
The probability of stopping at first traffic light is 0.6. If you stop at one set the prob you have to stop at the next is 0.9. If you dont have to stop at one set then the prob you dont have to stop at next is 0.7.

Find prob that you stop at exactly one set of lights.

I thought it was (0.6*0.1*0.7) + (0.4*0.3*0.9) + (0.4*0.7*0.3), which is wrong. Where is the mistake? Thanks

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Construct a contingency tree (like that in the attachment).

Identify those terminal nodes that correspond to the specified cases (the heavy lines in the attachment for this problem).

Multiply the branch probabilities together along the path leading to each terminal node corresponding to a specified case.

Finaly add up the terminal node probabilities to get:

(0.6x0.1x0.3)+(0.4x0.7x0.1)+(0.4x0.3x0.7).

RonL

Thank you again CaptainBlack! Hope it didnt take you too long to do that diagram...