In a satisfaction survey, 11% of a tour guide’s customers said that the tour was

Question

In a satisfaction survey, 11% of a tour guide’s customers said that the tour was too short. However, 48% said the tour was great. Estimate the probability that the guide will have to read at least 5 surveys to find one that said the tour was too short. Then estimate the probability that the guide will have to read at least 5 surveys to find one saying the tour was great.

3. Probability of reading at least 5 surveys to find one that said the tour was too short:

4. Probability of reading at least 5 surveys to find one that said the tour was great:

Explanation / Answer

P(tour was short) = 0.11, P(Tour was not short) = 0.89

P(Tour was great) = 0.48, P(Tour was not great) = 0.52

3.

P(fifth survey says tour was short ) = 0.89*0.89*0.89*0.89*0.11 = 0.069

4.

P(fifth survey says tour was great) = 0.52*0.52*0.52*0.52*0.48 = 0.035

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.