Questions: About 11% of the population of a large country is nervous around strangers. If two people are randomly selected, what is the probability both are nervous around strangers? What is the probability at least one is nervous around strangers? Assume the events are independent. (a) The probability that both will be nervous around strangers is .0121 (Round to four decimal places as needed.) (b) The probability that at least one person is nervous around strangers is (Round to four decimal places as needed.)

Solution Steps

To find the probability that both people are nervous around strangers, we use the binomial probability formula:

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

where:

• \( n = 2 \) (number of trials),
• \( x = 2 \) (number of successes),
• \( p = 0.11 \) (probability of success),
• \( q = 1 - p = 0.89 \) (probability of failure).

Substituting these values into the formula, we get:

\[ P(X = 2) = \binom{2}{2} \cdot (0.11)^2 \cdot (0.89)^0 = 1 \cdot 0.0121 \cdot 1 = 0.0121 \]

Thus, the probability that both people are nervous around strangers is \( 0.0121 \).

The probability that at least one person is nervous is the complement of the probability that none are nervous. Therefore, we first calculate the probability that none are nervous:

\[ P(X = 0) = \binom{2}{0} \cdot (0.11)^0 \cdot (0.89)^2 = 1 \cdot 1 \cdot 0.7921 = 0.7921 \]

The probability that at least one person is nervous is:

\[ 1 - P(X = 0) = 1 - 0.7921 = 0.2079 \]

Final Answer