Question

AHarris 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 (USA Today, January 21, 2004). a. In a sample of six international travelers, what is the probability that two will stick with their tour group? b. In a sample of six international travelers, what is the probability that at least two will stick with their tour group? c. In a sample of 10 international travelers, what is the probability that none will stick with the tour group?

Answer 1

Answer:

a) There is a 27.89% probability that two will stick with their tour group.

b) There is a 41.81% that in at least two will stick with their tour group.

c) There is a 7.33% probability that none will stick with their tour group.

Step-by-step explanation:

For each responders, this question can only have two outcomes. Either they stick with their tour group, or they don't. So this means that we can solve this problem as binomial distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

For these problems

The survey found that 23% of the respondents stick with their tour group, so $\pi = 0.23$.

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

The sample has six travelers, so $n = 6$.

We want P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

There is a 27.89% probability that two will stick with their tour group.

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

Either less than two stick with their groups, or at least two do. The sum of these probabilities is decimal 1.

So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

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$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

There is a 41.81% that in at least two will stick with their tour group.

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

Now our sample has 10 travelers, so $n = 10$.

We want to find P(X = 0)

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.23)^{0}.(0.77)^{10} = 0.0733$

There is a 7.33% probability that none will stick with their tour group.