The television show Degenerate Housewives has been successful for many years.

Question

That show recently had a share of 18, meaning that among the TV sets in use, 18% were tuned to Degenerate Housewives. Assume that an advertiser wants to verify that 18% share value by conducting its own survey, and a pilot survey begins with 10 households have TV sets in use at the time of a Degenerate Housewives broadcast.

Find the probability that none of the households are tuned to Degenerate Housewives.

P(none) = 

Find the probability that at least one household is tuned to Degenerate Housewives.

P(at least one) =

Find the probability that at most one household is tuned to Degenerate Housewives.

P(at most one) =

If at most one household is tuned to Degenerate Housewives, does it appear that the 18% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Degenerate Housewives unusual?)

1. no, it is not wrong
2. yes, it is wrong

Answer 1

Hi Anchal,

This is a binomial case since we have a binary outcome of "success" (tuned in) or "failure" (not). Formula is:

P(X=x)=C(n, x) p^x*q^n-x

Breaking this down:

X=number of viewers we are given to test

C(n, x)= n!/(x!(n-x)!)--exclamation point means factorial--i.e. 4!= 4*3*2*1--input this in calculator

p=probability viewer is tuned in

q=1-p

n=sample size

For your first problem:

X=0

p=0.18

q=0.82

n=10

Thus:

P(X=10)=C(10,0)*0.18⁰*0.82¹⁰

C(10,0)=10!/(0!*10!)=1

Additionally, anything raised to the 0 power is 1, so:

P(X=0)=1*1*0.82¹⁰

P(X=0)=0.137

You were then asked for the probability of at least one household being tuned in: you can use the Complement Rule aka the "One Minus Trick."

P(X>=1)=1-P(X=0)

P(X>=1)=1-0.137=0.863

Finally, you were asked to find the probability that at most one household tuned in. We already calculated the probability for 0, so we now need the probability for 1. All values are the same as in above equation except: x=1. Thus:

P(X=1)=C(10,1)*0.18¹*0.82⁹

C(10,1)=10!/(1!*9!)=10

P(X=1)=10*0.18¹*0.82⁹=0.302

Now, recall the question asked for at most 1, which means we also have to account for 0, so:

P(X<=1) = P(X=0) + P(X=1)

P(X<=1)= 0.302 + 0.137 = 0.439

This means that the occurrence of "At most 1" household tuned in happens about 43.9% of the time; not particularly unusual, so I would say the share value is not "wrong." Your instructor may use different criteria for this part, though. I hope this helps.