Math Stats probability problem!?

Question

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.

In a sample of six international travelers, what is the probability that at least two will stick with their group?!?

How is this solved?

Answer 1

This is a binomial distribution.

Let X be the number of travelers that stick with their tour group out of 6

X ~ Bin(6,0.23)

What you want is AT LEAST two, so the probability function is:

P(X>=2) = 1 -[ P(X=0) + P(X=1) ]

= 1 -[ 6C0 x 0.23^0 x 0.77^6 + 6C1 x 0.23^1 x 0.77^5 ]
= 1 - 0.58196
=0.418 (Ans)

Answer 2

Assume that the probability for an individual to stick with her tour group is 0.23. That means the probability that one will not is 0.77 and the situation is "binomial" .

The probability that none of the six sticks with her group is:
(0.77)**6.

The probability that exactly one sticks with the group is (0.23)*().77)**5 but that is the probability of a specific sequence and the order of the sequence is irrelevant so we must multiply this by the number of 6 things taken one at a time = ( 6, 1 ) = 6! (1 (5!) = 6.

The probability of there being two or more is (1-Probability that there is 0 - Probability that there is one ). =
1 -(0.77)**6 - 6 { 0.23)(0.77)**5 = 1 - 0.1605 - 0.2876 = 0.5519