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Secondary V • 3mo.

A police officer is using a radar device to check motorists' speeds. Prior to beginning the speed check, the officer estimates that 35 percent of motorists will be driving more than 5 miles per hour over the speed limit. Assuming that the police officer's estimate is correct, what is the probability that among 3 randomly selected motorists, the officer will find at least 1 motorist driving more than 5 miles per hour over the speed limit?

Mathematics
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Explanations (1)

  • Explanation from Alloprof

    Explanation from Alloprof

    This Explanation was submitted by a member of the Alloprof team.

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    Team Alloprof • 3mo.

    Thanks for your question!


    Let's establish what we know:

    • 35% of motorists drive over 5 miles per hour over the speed limit. By extension, the probability of drawing a motorist that is going over 5 miles per hour over the speed limit is 35%.

    • There are 3 randomly selected motorists

    We are looking for:

    • The probability of at least one of the randomly selected motorists driving more than 5 miles per hour over the speed limit


    To answer this question, let's think of all the possible ways in which at least one of the drivers is driving more than 5 miles per hour over the speed limit. There are 3 :

    1. 1 driver is over the limit and the 2 others are lawful drivers

    2. 2 drivers are over the limit and the 1 other is a lawful driver.

    3. All 3 drivers are over the limit.


    The probability of scenario 1 is the following:

    $$ P(speeding)•P(law\:abiding)•P(law\:abiding) $$

    $$ = 0.35•(1-0.35)•(1-0.35) = 0.252875 $$

    The probability of scenario 2 is the following:

    $$ P(speeding)•P(speeding)•P(law\:abiding) $$

    $$ = 0.35•0.35•(1-0.35) = 0.104125 $$

    The probability of scenario 3 is the following:

    $$ P(speeding)•P(speeding)•P(speeding) $$

    $$ = 0.35•0.35•0.35 = 0.042875 $$


    The cumulative probability of all three scenarios is thus the sum of their three probabilities:

    $$ P(\text{number of speeding drivers >=1}) = P(1)+P(2)+P(3) $$

    $$ P(\text{number of speeding drivers >=1}) = 0.252875+0.104125+0.042875 $$


    A quick tip: it is often useful to think of these scenarios using the words AND as well as OR. Any probability involving the word AND is a multiplication, whereas any probability involving the world OR is an addition. For instance:

    1 driver is speeding AND 1 driver is not speeding = P(speeding)xP(not speeding)

    1 driver is speeding OR 1 driver is not speeding =

    P(speeding) + P(not speeding)

    1 driver is speeding AND 1 is not speeding OR both drivers are not speeding =

    P(speeding)xP(not speeding) + P(not speeding)xP(not speeding)


    This webpage from the Alloprof website explains probabilities:


    Don't hesitate if you need more help!

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