An electrical engineer has on hand two boxes of resistors, with four resistors i

Question

An electrical engineer has on hand two boxes of resistors, with four resistors in each box. The resistors in the first box are labeled 9 Ohm, but in fact their resistances are 8, 9, 10, and 11 Ohm. The resistors in the second box are labeled 20 Ohm, but in fact their resistances are 18, 19, 20, and 21 Ohm. The engineer chooses one resistor from each box, and determines the resistance of each. Let A be the event that the first resistor has a resistance greater than 9, let B be the event that the second resistor has a resistance less than 19, and let C be the event that the sum of the resistances is equal to 27. Find B C and A B^c. B U C = {(8, 18), (10, 18), (11, 18), (8, 19)} A B^c = {(8, 19), (8, 20), (8, 21), (9, 19), (9, 20), (9, 21)} B U C = {(8, 18), (10, 18), (11, 18), (8, 19)} A B^c = {(10, 19), (10, 20), (10, 21), (11, 19), (11, 20), (11, 21)} B C = {(8, 18), (9, 18), (10, 18), (11, 18), (8, 19)} A B^c = {(10, 19), (10, 20), (10, 21), (11, 19), (11, 20), (11, 21)} B C = {(8, 18), (9, 18), (10, 18), (11, 18), (8, 19)} A B^c= {(8, 19), (8, 20), (8, 21), (9, 19), (9, 20), (9, 21)}

Explanation / Answer

A = {(10,18) (10,19) (10,20) (10,21) (11,18) (11,19) (11,20) (11,21)}

B = {(8,18) (9,18) (10,18) (11,18)}

C = {(9,18) (8,19)}

B or C =  {(8,18) (9,18) (10,18) (11,18) (8,19)}

A and B' = {(10,19) (10,20) (10,21) (11,19) (11,20) (11,21)}

Hence option (C) is correct

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