Naturalization. The U.S Bureau of citizenship and Immigration services collects and reports information about naturalized persons in Statistical Yearbook. Following is an age distribution for persons naturalized during one year.

Suppose that one of these naturalized person is selected at random.

Part (a) Without using the general addition rule, determine the probability that the age of the person obtained is either between 30 and 64. inclusive or at least 50.

Part (b) Find the probability in part (a), using the general rule.

Part (c) Which method did you find easier?

Age	Frequency	Age	frequency
18 - 19	5,958	45 - 49	42,820
20 - 24	50,905	50 - 54	32,574
25 -29	58,829	55 - 59	25,534
30 - 34	64,735	60 - 64	18,767
35 - 39	69,844	65 - 74	25,528
40 - 44	57,834	75 & over	9,872

The following table shows the age distribution of naturalised citizens over the course of a year:

Let E represent the possibility that the individual acquired is between the ages of 30 and 64, inclusive, or at least 50.

The total number of outcomes was 4,63,200.

In the case of a group of people older than 30,

The number of favourable results = 347508

$$\begin{gathered} P\left(E\right)=\frac{347508}{463200} \\ P\left(E\right)\approx 0.75 \end{gathered}$$

Let be the case where the individual's age is between 30 - 64

is the case where the individual's is at least 50.
$$\begin{gathered} E_1=64,735+69,844+57,834,+42,820+32,574,+25,534+18,767 \\ =312,108 \\ {E}_2=32,574+25,534+18,767+25,528+9872=112,275 \\ (E_1\cap E_2)=32,574+25,534+18,767=76,875 \\ P\left(E_1\right)=\frac{312108}{463200} \\ P\left(E_2\right)=\frac{112275}{463200} \\ P(E_1\cap E_2)=\frac{76875}{463200} \end{gathered}$$

Now, using general addition rule:

$$\begin{gathered} P(E_{1}or{E}_2)=P\left(E_1\right)+P\left(E_2\right)-P(E_1\cap E_2) \\ P(E_{1}or{E}_2)=\frac{347508}{463200} \end{gathered}$$

The first method is simpler since it requires less computation