Question

among 51 - to 56 - year - olds, 29% say they have written an editorial letter while under the influence of alcohol. suppose seven 51 - to 56 - year - olds are selected at random. complete parts (a) through (d) below.
0.0002
(round to four decimal places as needed.)
(b) what is the probability that at least one has not written an editorial letter while under the influence of alcohol?
0.9998
(round to four decimal places as needed.)
(c) what is the probability that none of the seven have written an editorial letter while under the influence of alcohol?
0.091
(round to four decimal places as needed.)
(d) what is the probability that at least one has written an editorial letter while under the influence of alcohol?
(round to four decimal places as needed.)

Explanation:

Step1: Identify probabilities

The probability that a 51 - to 56 - year - old has written an editorial letter while under the influence of alcohol is $p = 0.29$, so the probability that a 51 - to 56 - year - old has not written such a letter is $q=1 - p=1 - 0.29 = 0.71$. We use the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here $n = 7$.

Step2: Calculate probability for part (d)

The probability that at least one has written an editorial letter while under the influence of alcohol is the complement of the event that none of them has written such a letter.
The probability that none of them has written such a letter is $P(X = 0)=C(7,0)\times(0.29)^{0}\times(0.71)^{7}$.
Since $C(7,0)=\frac{7!}{0!(7 - 0)!}=1$ and $(0.29)^{0}=1$, then $P(X = 0)=1\times1\times(0.71)^{7}\approx0.091$.
The probability that at least one has written an editorial letter while under the influence of alcohol is $P(X\geq1)=1 - P(X = 0)=1-(0.71)^{7}\approx1 - 0.091 = 0.9090$.

Answer:

$0.9090$