Question

the probability of catching the flu this year is 0.15. use the binomial model to find the probability that 2 out 5 members of the dawson family get the flu this year.

$p(x) = \left \frac{n!}{x!(n - x)!} \
ight p^x q^{n - x}$

(1 point)

\\(\bigcirc\\) 2.43%
\\(\bigcirc\\) 10.7%
\\(\bigcirc\\) 13.8%
\\(\bigcirc\\) 23.7%

Explanation:

Step1: Identify values

Here, \( n = 5 \) (number of trials), \( x = 2 \) (number of successes), \( p = 0.15 \) (probability of success), \( q = 1 - p = 1 - 0.15 = 0.85 \) (probability of failure).

Step2: Calculate combination

First, calculate the combination \( \frac{n!}{x!(n - x)!} \). Substituting values: \( \frac{5!}{2!(5 - 2)!} = \frac{5!}{2!3!} = \frac{5\times4\times3!}{2\times1\times3!} = 10 \).

Step3: Calculate \( p^x \) and \( q^{n - x} \)

\( p^x = 0.15^2 = 0.0225 \), \( q^{n - x} = 0.85^{3} = 0.85\times0.85\times0.85 = 0.614125 \).

Step4: Multiply all parts

Now, multiply the combination, \( p^x \), and \( q^{n - x} \): \( P(2) = 10\times0.0225\times0.614125 = 10\times0.0138178125 = 0.138178125 \). Convert to percentage: \( 0.138178125\times100\% \approx 13.8\% \).

Answer:

13.8% (corresponding to the option with 13.8%)