Question

some bags of rice containing the following weights of the rice ( in Kg): 4.97, 3.999, 5.05,5.08,5.03,5.00,4.35,5.06, 5.00,5,07,5.04,5.00,4.098,5.001. find the probability that any one of these bags chosen at random contains less than 5Kg of rice

Answer 1

Hi ,

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The probability of an event is written as P (E )  and defined as ,

P ( E ) = ( Number of outcomes favorable to E ) / ( Number of all possible  

                                                           outcome of the experiment )

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According to the given problem ,

Total possible outcomes are 4.97 , 3.99 , 5.05 , 5.08 , 5.03 , 5.00 , 4.35 , 

5.06 , 5.00 , 5.07 , 5.04 , 5.00 , 4.098 , 5 .001 

Number of total possible outcomes = 14 --------( 1 )

Favorable outcome = bags choosen at random contain less than 

                                    5kg rice 

                                = 4.97 , 3.999 , 4.35 , 4.098

Number of favorable outcomes = 4 --------------------- ( 2 )
 
 P ( bags containing less than 5 kgs rice ) = ( 2 ) / ( 1 )

                                                                   = 4  / 14  

                                                                   = 2 / 7   

I hope this helps you .
 
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Answer 2

Answer:

Total number of bags = 15Total number of possible outcomes = 15Number of bags containing rice less than 5 kg = 4Number of favourable outcomes = 4P(selecting a bag containing rice less than 5 kg) = Number of favourable outcomes Total number of possible outcome=4/15