Question

The television show Lett3rs has been successful for many years. That show recently had a share of 24, meaning that among the TV sets in use, 24% were tuned to Lett3rs. Assume that an advertiser wants to verify that 24% share value by conducting its own survey, and a pilot survey begins with 11 households have TV sets in use at the time of a Lett3rs broadcast.

Find the probability that none of the households are tuned to Lett3rs. P(none) =
Find the probability that at least one household is tuned to Lett3rs. P(at least one) =
Find the probability that at most one household is tuned to Lett3rs. P(at most one) =
If at most one household is tuned to Lett3rs, does it appear that the 24% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Lett3rs unusual?)

Answer 1

(a) P(none) = 0.0488

(b) P(at least one) = 0.9512

(c) P(at most one) = 0.2186

(d) No, it doesn't appear that the 24% share value is wrong.

Explanation:

We are given that among the TV sets in use, 24% were tuned to Lett3rs.

A pilot survey begins with 11 households have TV sets in use at the time of a Lett3rs broadcast.

Let X = Number of households that are tuned to Lett3rs

The above situation can be represented through binomial distribution;

`P(X = r)=\binom{n}{r}* p^(r) * (1-p)^(n-r) ; x = 0,1,2,3,......`

where, n = number of samples (trials) taken = 11 households

r = number of success

p = probability of success which in our question is % of

households that were tuned to Lett3rs, i.e; 24%

So, X ~ Binom{n = 11, p = 0.24}

(a) The probability that none of the households are tuned to Lett3rs is given by = P(X = 0)

P(X = 0) =
`\binom{11}{0}* 0.24^(0) * (1-0.24)^(11-0)`

=
`1* 1 * 0.76^(11)`

= 0.0488

(b) The probability that at least one households are tuned to Lett3rs is given by = P(X
`\geq` 1)

P(X
`\geq` 1) = 1 - P(X = 0)

= 1 - 0.0488

= 0.9512

(b) The probability that at most one households are tuned to Lett3rs is given by = P(X
`\leq` 1)

P(X
`\leq` 1) = P(X = 0) + P(X = 1)

=
`\binom{11}{0}* 0.24^(0) * (1-0.24)^(11-0)+\binom{11}{1}* 0.24^(1) * (1-0.24)^(11-1)`

=
`1* 1* 0.76^(11)+11* 0.24^(1) * 0.76^(10)`

= 0.2186

No, it doesn't appear that the 24% share value is wrong because the probability that at most one household is tuned to Lett3rs is somewhere close to 24% and this probability is also not unusual as it is way more than the criteria of more than 5%.