Question

# Census reports for a city indicate that 62% of residents classify themselves as Christian, 12% as Jewish , and 16% as members of other religions (Musl

Probability

Census reports for a city indicate that $$62\%$$ of residents classify themselves as Christian, $$12\%$$ as Jewish , and $$16\%$$ as members of other religions (Muslims, Buddhists, etc.).
The remaining residents classify themselves as nonreligious.
A polling organization seeking information about public opinions wants to be sure to talk with people holding a variety of religious views, and makes random phone calls. Among the first four people they call, what is the probability they reach

a) all Christians?

b) no Jews?

c) at least one person who is nonreligious?

2021-02-27

A) Use the multiplication rule
$$P(all christians)= 62% xx 62% xx 62% xx 62%=0.62^4 approx 0.1478=14.78%$$
B) The sum of all probabilities should be $$100\%$$:
P(not a jew) $$= 100\% - 12\% =88\%$$
Use the multiplication rule
$$P(no jews)=88% * 88% * 88% * 88% = 0.88^4 approx 0.5997=59.97%$$
C)Use the addition rule
P(religious) $$= 62\%+12\%+16\%=90\%$$
P(at least one non religious)$$=100\% - P$$(all religious) $$= 100%-90% * 90% * 90% * 90% approx 0.3439 =34.39%$$
Result:
(a) $$14.78\%$$
(b) $$59.97\%$$
(c) $$34.39\%$$