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A metal gate of dimension 5 m × 1.5 m (width* height) has some ‘n’ number of division each of equal dimension of 5 cm × 2 cm (width* height). One division is selected from this ‘n’ division.
Quantity I: Probability that the selected division is in the topmost row
Quantity II: Probability that the selected division in the leftmost column
Quantity I > Quantity II
Quantity I ≥ Quantity II
Quantity II > Quantity I
Quantity II ≥ Quantity I
Quantity I = Quantity II or Relation cannot be established
Number of column= 500/5=100 Number of rows= 150/2=75 total division=100*75 I: P (top row)= top row division/ total division= 100/75*100= 1/75 II: P(left most column)= left row division/total division= 75/75*100=1/100
Hence, option 1 is the correct answer.
By: Amit Kumar ProfileResourcesReport error
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