Suppose that in a year, 10% of all light vehicles (SUVs, pickups, passenger cars, and minivans) sold in the United States were pickups and 30% were passenger cars. Moreover, a randomly chosen vehicle sold that year was twice as likely to be an SUV as a minivan. Find the associated probability distribution.
Outcome | SUVs | pickups | passenger cars | minivans |
Probability |
SOLUTION:
From given data,
Suppose that in a year, 10% of all light vehicles (SUVs, pickups, passenger cars, and minivans) sold in the United States were pickups and 30% were passenger cars. Moreover, a randomly chosen vehicle sold that year was twice as likely to be an SUV as a minivan. Find the associated probability distribution.
Let ,
P(SUV) = 2x
P(minivan) = x = 0.2(from below calculation)
P(pickup) = 10% = 10/100 = 0.1
P(passenger car) = 30% = 30 / 100 =0.3
Therefore,
P(SUV) +P(pickup) + P(passenger car) +P(minivan) = 1
2x + 10/100 + 30 / 100 + x = 1
3x + 40 / 100 =1
3x = 1 - ( 40 / 100 )
3x = (100 -40) /100
3x = 60 /100
x = 20 / 100
x = 0.2
P(SUV) = 2x = 2*0.2 = 0.4
Outcome | SUVs | pickups | passenger cars | minivans |
Probability | 0.4 | 0.1 | 0.3 | 0.2 |
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