129k views
3 votes
Hi, can you help me answer this question, please, thank you:)

Hi, can you help me answer this question, please, thank you:)-example-1

1 Answer

4 votes

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;


\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}

(b) The probability that at least one household is tuned to lett3rs is;


\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}

(c) The probability that at most one household is tuned to lett3rs is;


\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}

(d) Since the probability of at most one is less than 5%, then No, It is not wrong.

User Heston Liebowitz
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.