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.
(a) what is the probability that all seven have written an editorial letter while under the influence of alcohol?
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?
(round to four decimal places as needed.)

Explanation:

Step1: Identify probability of success and number of trials

Let $p = 0.29$ be the probability that a 51 - to 56 - year - old has written an editorial letter while under the influence of alcohol, and $n=7$ be the number of people selected.

Step2: Calculate probability for part (a)

The probability that all $n = 7$ have written an editorial letter while under the influence of alcohol is given by the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$. When $k = 7$, $C(7,7)=1$, so $P(X = 7)=p^{7}=(0.29)^{7}\approx0.0002$.

Step3: Calculate probability for part (b)

The probability that at least one has not written an editorial letter while under the influence of alcohol is the complement of the event that all have written such a letter. Let $A$ be the event that all have written an editorial letter while under the influence of alcohol. Then $P(\text{at least one has not})=1 - P(A)$. Since $P(A)\approx0.0002$, then $P(\text{at least one has not})=1-(0.29)^{7}=1 - 0.0002 = 0.9998$.

Answer:

(a) $0.0002$
(b) $0.9998$