Question

A recent survey found that 74% of households had Internet access and 70% of households had cable television. Also, it was reported that 69% 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 P(Internet or cable)- Type an integer or a decimal) (u) More

Answer 1

P(Internet or cable) =75%

The correct option is B) 0.9658

0.70 % = Given. Probability of household hod internet access PCA)= 747. Probability of household hod cable television PLB) = 70. Probability of house hold had both internet and able television PCAN B)= 69% =0.69 Now, we want to find the probability that house had either internet access or cable television ie PC A or B) = Pl Internet or cable) We know PLAUB)= PLA) + PCB) – PLAN B) :: PCA or B)= P(AUB) = PLA)+P1B)- PIAN B) = 0.74 +0.70 -0.69 = 0.75 P(Internet or cable) = 75% Let X denote the number of chocolate chips per cookie follow the poisson distribution with porometer d=5.2 jie XN PLd=5.2) The probability mass function of xis PC X= n] = e-?_ 5.25.2 Red 200, 1, 2, --. - H! x! Now. Probability that a rondomly selected cookie will contain two or more chocolate chips. iie. PCX > 2] = 1-PCX22]. = 1-[p[x=0] +P[X= 137 - - [-9.2.520, esse] 51-[0.00652 +0.02869] 50.96579 P[ X7, 2] = 0.9658 The correct option is ® 0.9658