in a survey of 500 families number of vehicles owned per family we found to be as followed number of vehicles 0 1 2 3 4 number of families 35 213 170 65 17 find the probability that a family selected add random as a) no vehicals b) 2 or 3 vehicels c) more than 1 vehicals

@Prachet, Good work!!@Sakina, your solution is as below,a) Number of families with no vehicle = 35Probability = Number of families with no vehicleTotal Number of families=35500=7100b) Number of families with 2 or 3 vehicles = 35Probability = Number of families with 2 or 3 vehiclesTotal Number of families=170+65500=235500=47100c) Number of families with more than 1 vehicles = 35Probability = Number of families withmore than 1 vehiclesTotal Number of families=170+65+17500=252500=64125

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a)- probability(no vehicles)= 35/500=7/100

b)probability(2 or 3 vehicles)=235/500=47/100

c)- probability (more than 1 vehicle)=253/500
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