prob. two of six travellers stay with tour group, etc.

A Harris Interactive survey for InterContinental Hotels & 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.

a. In a sample of 6 international travelers, what is the probability that two will stick with their tour group?

b. In a sample of 6 international travelers, what is the probability that at least two will stick with their tour group?

c. In a sample of 10 international traveleers, what is the probability that none will stick with their tour group?

I don't even know where to begin?? any help will be appreciated.

Thanks!

Where to begin is looking at the binomial probability.

\(\displaystyle \L\\C(n,k)p^{k}q^{n-k}\)

q=1-p.

Here's how to set up the first one.

\(\displaystyle \L\\C(6,2)(0.23)^{2}(0.77)^{4}\)

If they say 'at least 2', then you can sum up from 2 to the number in the sample, in this case 6.

\(\displaystyle \L\\\sum_{k=2}^{6}C(6,k)(0.23)^{k}(0.77)^{6-k}\)

or equivalently, you can find the probabilitiy of at most 1 and subtract from 1.

\(\displaystyle \L\\1-\sum_{k=0}^{1}C(6,k)(0.23)^{k}(0.77)^{6-k}\)

For none, set it up as in the first example, only use k=0 and n=10.