Question

according to a poll, 30% of voters support a ballot initiative. hans randomly surveys 5 voters. what is the probability that exactly 2 voters will be in favor of the ballot initiative? round the answer to the nearest thousandth.
\\( p(k \text{ successes}) = {}_nc_k p^k (1 - p)^{n - k} \\)
\\( {}_nc_k = \frac{n!}{(n - k)! k!} \\)
\\( \circ 0.024 \\)
\\( \circ 0.031 \\)
\\( \circ 0.132 \\)
\\( \circ 0.309 \\)

Explanation:

Step1: Identify values

Here, \( n = 5 \) (number of voters), \( k = 2 \) (successes), \( p = 0.3 \) (probability of support).

Step2: Calculate combination \( _nC_k \)

Using \( _nC_k=\frac{n!}{(n - k)!k!} \), substitute \( n = 5 \), \( k = 2 \):

$$ _5C_2=\frac{5!}{(5 - 2)!2!}=\frac{5\times4\times3!}{3!\times2\times1}=\frac{20}{2}=10 $$

Step3: Calculate probability formula

Using \( P(k\text{ successes}) = _nC_kp^k(1 - p)^{n - k} \), substitute values:

$$ P(2)=10\times(0.3)^2\times(1 - 0.3)^{5 - 2}=10\times0.09\times(0.7)^3 $$

Step4: Compute powers and multiply

\( (0.7)^3 = 0.343 \), so:

$$ P(2)=10\times0.09\times0.343 = 10\times0.03087 = 0.3087\approx0.309 $$

Wait, no, wait. Wait, \( (0.3)^2=0.09 \), \( (0.7)^3 = 0.343 \), \( 10\times0.09 = 0.9 \), \( 0.9\times0.343 = 0.3087 \approx 0.309 \)? But wait, maybe I miscalculated. Wait, no, let's recheck:

Wait, \( _5C_2 = 10 \), \( p = 0.3 \), \( k = 2 \), \( (1 - p)=0.7 \), \( n - k = 3 \).

So \( P(2)=10\times(0.3)^2\times(0.7)^3 = 10\times0.09\times0.343 = 10\times0.03087 = 0.3087 \approx 0.309 \). But the options have 0.309 as an option. Wait, but maybe I made a mistake? Wait, no, let's check again.

Wait, \( (0.3)^2 = 0.09 \), \( (0.7)^3 = 0.343 \), multiply 0.09 and 0.343: 0.09 * 0.343 = 0.03087, then multiply by 10: 0.3087, which rounds to 0.309. So the correct answer is 0.309.

Answer:

0.309