Question 78623: On your way to a particular location that you regularly travel, There are three intersections with traffic lights. Each of the three lights has a different stop and go pattern: Light one is green 10% of the time, light two is green 40% of the time, and light three is green 25% of the time. Assuming that you are a very safe driver and only drive through an intersection when the light is green, what is the probability that you will arrive at your destination without having to stop(that is,hitting all three traffic lights green)?
a 1%
b 25%
c 75%
d 100%
e none of the above
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! With independent events such as these, the probabilities of success (lights green) just
multiply.
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When you get to the first light the chance that it will be green is 10% (0.1 or 1 out of 10)
and that it is other than green is 90% (0.9 or 9 out of 10). Suppose that it is green and
you go directly to the second light.
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The odds of the second light being green is 40% (0.4 or 4 out of 10). If you get through
it on green you have cleared two lights ... a 0.1 chance of the first light and a 0.4 chance
on the second light. The chances of this combination is 0.1 times 0.4 which is 0.04 or 4%.
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Finally, you approach the third light in the sequence. It has a 25% (0.25 or 1 out of
4) chance of being green. This probability multiplies the probability that the previous
combination was green (0.04). So the overall probability of all three events being "green
light" is 0.04 times 0.25 which is 0.01 or 1%.
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In summary, the chances of all three being green are: 0.1*0.4*0.25 = 0.01 or 1%.
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Let's expand the problem a little. What are the chances that you will clear the first light,
miss the second light, and then clear the third light? You work this problem the same
general way, by multiplying the probabilities of each individual event. Clearing the
first light is 10% (0.1), missing the second light is 60% probability (since making it
was 40% you subtract 40% from 100% to find that the chance of missing it is 60%) and finally
making the third light is 25%. So the probability of make - miss - make is:
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0.1 * 0.6 * 0.25 = .015 or 1.5%
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Similarly, the probability of missing all three lights involves 90% (0.9) on the first light,
60% (0.6) on the second light, and 75% on the third light and this translates to:
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0.9*0.6*0.75 = 0.405 or 40.5%
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Try this one for yourself ... what is the probability that you make the first two lights
but miss the third one. You should get an answer of 3%.
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There are 8 possible outcomes as follows:
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make - make - make <=== from above 1%
make - make - miss <=== from above 3%
miss - make - make
miss - make - miss
make - miss - make <=== from above 1.5%
make - miss - miss
miss - miss - make
miss - miss - miss <=== from above 40.5%
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If you calculate the probability of each of these combinations and add up your answers,
the total should be 100% because there is a 100% chance that one of these sequences
of lights will happen.
.
Hope this helps you to understand this type of probability a little better.
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