Question

In a small city, approximately 14% of those eligible are called for jury duty in any one calendar year. People are selected for jury duty at random from those eligible, and the same individual cannot be called more than once in the same year. (a) What is the probability that a particular eligible person in this city is selected in both of the next 2 years? (Enter your answer to four decimal places.) (b) What is the probability that a particular eligible person in this city is selected in all three of the next 3 years? (Enter your answer to six decimal places.)

Answer 1

Answer:

a

  $P(E \ n\ Z) = 0.0196$

b

$P(E \ n\ Z \ n \ Y) = 0.0027$

Step-by-step explanation:

From the question we are told that

The probability of that a person is being called for jury duty in any given year is $P(J) = 0.14$

Let P(E) be the probability a person being selected in the first year, Let P(Z) be the probability a person being selected in the second year

and Let P(Y) be the probability a person being selected in the third year

Generally the probability that a particular eligible person in this city is selected in both of the next 2 years is mathematically represented as

$P(E \ n\ Z) = P(E) * P(Z) = P(J)^2$

=>      $P(E \ n\ Z) = 0.14^2$

=>      $P(E \ n\ Z) = 0.0196$

Generally the  probability that a particular eligible person in this city is selected in both of the next 3 years is mathematically represented as

     $P(E \ n\ Z \ n \ Y) = P(E) * P(Z) * P(Y) = P(J)^3$

=> $P(E \ n\ Z \ n \ Y) = 0.14^3$

=>  $P(E \ n\ Z \ n \ Y) = 0.0027$

This multiplication of each probabilities is valid because each probability is independent of one another