Question

a wheel has 10 equally sized slices numbered from 1 to 10.
some are grey and some are white.
the slices numbered 1, 3, 5, 7, 8, 9, and 10 are grey.
the slices numbered 2, 4, and 6 are white.
the wheel is spun and stops on a slice at random.
let ( x ) be the event that the wheel stops on a white slice, and let ( p(x) ) be the
probability of ( x ).
let not ( x ) be the event that the wheel stops on a slice that is not white, and let
( p(\text{not } x) ) be the probability of not ( x ).
(a) for each event in the table, check the outcome(s) that are contained in the event. then, in the last column, enter the probability of the event.

(b) subtract.

Explanation:

Step1: Determine the number of favorable outcomes for \(X\)

Event \(X\) is the wheel stopping on a white slice. The white - numbered slices are \(2\), \(4\), and \(6\). So, the number of favorable outcomes for \(X\) is \(n(X)=3\).
The total number of slices \(n = 10\).
Using the probability formula \(P(X)=\frac{n(X)}{n}\), we have \(P(X)=\frac{3}{10}\).

Step2: Determine the number of favorable outcomes for \(not\ X\)

Event \(not\ X\) is the wheel stopping on a non - white (grey) slice. The grey - numbered slices are \(1\), \(3\), \(5\), \(7\), \(8\), \(9\), and \(10\). So, the number of favorable outcomes for \(not\ X\) is \(n(not\ X)=7\).
Using the probability formula \(P(not\ X)=\frac{n(not\ X)}{n}\), we have \(P(not\ X)=\frac{7}{10}\).

Step3: Calculate \(1 - P(not\ X)\)

Substitute \(P(not\ X)=\frac{7}{10}\) into \(1 - P(not\ X)\).

$$1-\frac{7}{10}=\frac{10 - 7}{10}=\frac{3}{10}$$

Answer:

• For event \(X\): Check the boxes for \(2\), \(4\), \(6\); \(P(X)=\frac{3}{10}\)
• For event \(not\ X\): Check the boxes for \(1\), \(3\), \(5\), \(7\), \(8\), \(9\), \(10\); \(P(not\ X)=\frac{7}{10}\)
• For \(1 - P(not\ X)\): \(\frac{3}{10}\)