Questions: Arachnophobia: A 2005 Gallup Poll found that 7% of teenagers (ages 13 to 17) suffer from arachnophobia and are extremely afraid of spiders. At a summer camp, there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other. (a) Calculate the probability that at least one of them suffers from arachnophobia. (b) Calculate the probability that exactly 2 of them suffer from arachnophobia. (c) Calculate the probability that at most 1 of them suffers from arachnophobia. (d) If the camp counselor wants to make sure no more than 1 teenager in each tent is afraid of spiders, does it seem reasonable for him to randomly assign teenagers to tents?

Solution Steps

To find the probability that at least one teenager suffers from arachnophobia, we can use the complement rule:

\[ P(\text{at least one}) = 1 - P(\text{none}) \]

The probability of none suffering from arachnophobia is given by:

\[ P(X = 0) = \binom{n}{0} \cdot p^0 \cdot q^{n} = 1 \cdot 1 \cdot (0.93)^{10} \approx 0.484 \]

Thus, the probability that at least one suffers from arachnophobia is:

\[ P(\text{at least one}) = 1 - 0.484 = 0.5160 \]

To calculate the probability that exactly 2 teenagers suffer from arachnophobia, we use the binomial probability formula:

\[ P(X = 2) = \binom{n}{2} \cdot p^2 \cdot q^{n-2} \]

Calculating this gives:

\[ P(X = 2) = \binom{10}{2} \cdot (0.07)^2 \cdot (0.93)^{8} \approx 0.1234 \]

To find the probability that at most 1 teenager suffers from arachnophobia, we sum the probabilities of 0 and 1 suffering from arachnophobia:

\[ P(\text{at most 1}) = P(X = 0) + P(X = 1) \]

We already calculated \(P(X = 0) \approx 0.484\). Now we calculate \(P(X = 1)\):

\[ P(X = 1) = \binom{10}{1} \cdot (0.07)^1 \cdot (0.93)^{9} \approx 0.3643 \]

Thus,

\[ P(\text{at most 1}) = 0.484 + 0.3643 \approx 0.8483 \]

Final Answer