&nbsp;A Harris Interactive survey for InterContinental Hotels &amp; Resorts asked respondents, “When traveling internationally, do you generally venture out on your own to experience culture, or stick with your tour group and itineraries?” The survey found that 23% of the respondents stick with their tour group (USA Today, January 21, 2004).In a sample of six international travelers, what is the probability that two will stick with their tour group?In a sample of six international travelers, what is the probability that at least two will stick with their tour group?In a sample of 10 international travelers, what is the probability that none will stick with the tour group?

Question

&amp;nbsp;A Harris Interactive survey for InterContinental Hotels &amp;amp; Resorts asked respondents, “When traveling internationally, do you generally venture out on your own to experience culture, or stick with your tour group and itineraries?” The survey found that 23% of the respondents stick with their tour group (USA Today, January 21, 2004).In a sample of six international travelers, what is the probability that two will stick with their tour group?In a sample of six international travelers, what is the probability that at least two will stick with their tour group?In a sample of 10 international travelers, what is the probability that none will stick with the tour group?

nick1337 · Accepted Answer

1)&amp;nbsp;f(x)=n!x!(n-x)!px(1-p)n-xLet,&amp;nbsp;n&amp;nbsp;=&amp;nbsp;the amount of people sampled&amp;nbsp;x&amp;nbsp;=&amp;nbsp;the amount of people to leave group&amp;nbsp;p&amp;nbsp;=&amp;nbsp;the probability of&amp;nbsp;x&amp;nbsp;people leaving group in n sampledUsing the formula above, calculate the probability that two will stick with their tour group when six people were sampled.f(x)=6!2!(6-2)!0.232(1-0.23)6-2=(6)(5)(4)(3)(2)(1)(2×1)(4×3×2×1)(0.23)2(0.77)4=(72048)(0.0529)(0.3515)=0.2789The probability that two will stick with their tour group from a sample of six is&amp;nbsp;0.27892) In this scenario, if at least two will stick, then&amp;nbsp;4&amp;nbsp;or&amp;nbsp;5&amp;nbsp;may leave the group. The formula used is the same used in part (1) but for each possible member leaving. Therefore, each possible person to leave must be found and added to the previous amount. List the modified formula below.f(at&amp;nbsp;least&amp;nbsp;2)=f(2)+f(3)+f(4)+f(5)+f(6)Using the formula above, calculate the probability that at least two stick with the tour group.P(at&amp;nbsp;least&amp;nbsp;2)=f(2)+f(3)+f(4)+f(5)+f(6)f(x)=6!2!(6-2)!0.232(1-0.23)6-2+6!3!(6-3)!0.233(1-0.23)6-3+6!4!(6-4)!0.234(1-0.23)6-4+6!5!(6-5)!0.235(1-0.23)6-5+6!6!(6-6)!0.236(1-0.23)6-6=0.2789+0.1111+0.0249+0.0030+0.0001=0.4180Out of 6 sampled, the probability that at least two will stick with the group is&amp;nbsp;0.41803) Here, in this scenario,&amp;nbsp;10&amp;nbsp;travelers are sampled and none are expected to stick with the group. Using the same formula, let&amp;nbsp;n&amp;nbsp;=&amp;nbsp;10&amp;nbsp;x&amp;nbsp;=&amp;nbsp;0&amp;nbsp;and,&amp;nbsp;p&amp;nbsp;=&amp;nbsp;0.23&amp;nbsp;Calculate the probability using those numbersf(x)=10!0!(10-0)!0.230(1-0.23)10-0=(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)(0×1)(10×9×8×7×6×5×4×3×2×1)(0.23)0(0.77)10=(1)(0.0733)=0.0733Out of 10 sampled, the probability that none are expected to stay is&amp;nbsp;0.0733

A Harris Interactive survey for InterContinental Hotels &

Answered question

Answer & Explanation

New Questions in Trigonometry