Question

1. -/1 points details my notes ask your teacher practice another illowskyintrostat1 4.3.044.pr. the higher education research institute at ucla collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the u.s. 71.8% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. suppose that you randomly pick seven first-time, full-time freshmen from the survey. you are interested in the number that believes that same-sex couples should have the right to legal marital status. what is the probability that at least two of the freshmen reply \yes\? (round your answer to four decimal places.)

Explanation:

Step1: Define binomial parameters

We have a binomial distribution with $n=7$ (number of trials), $p=0.718$ (probability of success), and we want $P(X \geq 2)$.

Step2: Use complement rule

Calculate $1 - P(X=0) - P(X=1)$, since $P(X \geq 2) = 1 - P(X < 2)$.

Step3: Compute $P(X=0)$

Binomial formula: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
$P(X=0) = \binom{7}{0}(0.718)^0(1-0.718)^{7-0} = 1 \times 1 \times (0.282)^7 \approx 0.00016$

Step4: Compute $P(X=1)$

$P(X=1) = \binom{7}{1}(0.718)^1(0.282)^{6} = 7 \times 0.718 \times (0.282)^6 \approx 0.0029$

Step5: Calculate final probability

Subtract the two probabilities from 1:
$1 - 0.00016 - 0.0029 = 0.99694$

Answer:

0.9969