Given that 29% were tuned to lett3rs;
Then, 71% were not tuned to lett3rs.
(a) The probability that none of the households are tuned to lett3rs is;
[tex]\begin{gathered} P(\text{none)}=^{15}C_0(0.29)^0(0.71)^{15} \\ P(\text{none)}=1(1)(0.0059) \\ P(\text{none)}=0.0059 \end{gathered}[/tex]
(b) The probability that at least one household is tuned to lett3rs is;
[tex]\begin{gathered} P(at\text{ least one)=1-P(none)} \\ P(at\text{ least one)=1-0.0059} \\ P(at\text{ least one)=}0.9941 \end{gathered}[/tex]
(c) The probability that at most one household is tuned to lett3rs is;
[tex]\begin{gathered} P(at\text{ most one)=P(none)+P(one household)} \\ P(^{}onehousehold)=^{15}C_1(0.29)^1(0.71)^{14} \\ P(^{}onehousehold)=15(0.29)(0.0083) \\ P(^{}onehousehold)=0.0361 \\ P(at\text{ most one)=0.0059+0.0361} \\ P(at\text{ most one)=}0.0420 \end{gathered}[/tex]
(d) Since the probability of at most one is less than 5%, then No, It is not wrong.
