Question

A recent survey found that

74 %74%

of households had Internet access and

73 %73%

of households had cable television.​ Also, it was reported that

65 %65%

of the households in the survey had both Internet and cable television. Determine the probability of selecting a household that had either Internet access or cable television.

Answer 1

A: Event of selecting an household that had internet access

Given , 74% of households had Internet access.

Therefore

P(A) = 74/100 = 0.74

B: Event of an seleccting an household that had cable television

Given , 73% of households had cable television

Therefore

P(B) = 73/100 = 0.73

Given, 65% house holds had both Internet and cable television.

A and B = Event of selecting of an house that had both Internet and cable television

P(A and B) = 65/100 = 0.65

A or B = Event of selecting an house that had either Internet and cable television

Probability of selecting a household that had either Internet access or cable television = P( A or B)

By Addition theorem of probability,

P( A or B) = P(A) + P(B) - P(A and B) = 0.74 + 0.73 - 0.65 = 0.82

Probability of selecting a household that had either Internet access or cable television = 0.82