Q. 45 Roulette An American roulette wh... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 5: Q. 45 (page 328)

Roulette An American roulette wheel has 38 slots with numbers $1$ through $36, 0,$ and $00$ , as shown in the figure. Of the numbered slots, $18$ are red, $18$ are black, and $2$ 鈥攖丑别 $0$ and $00$ 鈥攁re green. When the wheel is spun, a metal ball is dropped onto the middle of the wheel. If the wheel is balanced, the ball is equally likely to settle in any of the numbered slots. Imagine spinning a fair wheel once. Define events $B$ : ball lands in a black slot, and $E$ : ball lands in an even-numbered slot. (Treat $0$ and $00$ as even numbers.)
a. Make a two-way table that displays the sample space in terms of events $B$ and $E$ .
b. Find $P (B)$ and $P (E)$ .
c. Describe the event 鈥� $B$ and $E$ 鈥� in words. Then find the probability of this event.
d. Explain why $P (B o r E) 鈮� P (B) + P (E)$ . Then use the general addition rule to compute $P (B o r E)$ .

Short Answer

Expert verified

a. A two way table displaying the sample space of events $B$ and $E$ .

Sample space of event $B$ = $2, 35, 4, 33, 6, 31, 8, 29, 10, 13, 24, 15, 22, 17, 20, 11, 26, 28$

Sample space of event $E$ = $0, 2, 4, 16, 6, 18, 8, 12, 10, 00, 36, 24, 34, 22, 32, 20, 30, 26, 28$

b. The probability of event $B$ = $0.4737$ and event $E$ = $0.5$

c. The probability of the event 鈥� $B$ and $E$ 鈥� = $0.2632$

d. The probability of the event 鈥� $B$ or $E$ 鈥� = $0.7105$

Step by step solution

Part (a) Step 1 : Given Information

We have to determine a two way table displaying the sample space of events $B$ and $E$ .

Part (a) Step 2 : Simplification

A roulette wheel has a total of

38

slots.

There are

18

red slots in all.

There are

18

black slots in all.

The number of green slots is two.

The event

B

is when a ball lands in a black slot.

The event

E

is when a ball lands in an even-numbered slot.

Sample space of event $B$ = $2, 35, 4, 33, 6, 31, 8, 29, 10, 13, 24, 15, 22, 17, 20, 11, 26, 28$

Sample space of event $E$ = $0, 2, 4, 16, 6, 18, 8, 12, 10, 00, 36, 24, 34, 22, 32, 20, 30, 26, 28$

Part (b) Step 1 : Given Information

We have to determine the probability of event $B$ and event $E$ .

Part (b) Step 2 : Simplification

We fill use the following formula :-

$P (B) = \frac{T o t a l n u m b e r o f o u t c o m e s i n t h e s a m p l e s p a c e i n t h e e v e n t B}{T o t a l n u m b e r o f o u t c o m e s} P (E) = \frac{T o t a l n u m b e r o f o u t c o m e s i n t h e s a m p l e s p a c e i n t h e e v e n t E}{T o t a l n u m b e r o f o u t c o m e s}$

$P (B) = \frac{18}{38} = 0.4737 P (E) = \frac{19}{38} = 0.5$

Part (c) Step 1 : Given Information

We have to determine the probability of the event 鈥� $B$ and $E$ 鈥�

Part (c) Step 2 : Simplification

We will use the following formula :-

$P (B 鈭� E) = \frac{T o t a l n u m b e r o f o u t c o m e s i n t h e s a m p l e s p a c e i n t h e e v e n t (B 鈭� E)}{T o t a l n u m b e r o f o u t c o m e s}$

The ball lands in the black even-numbered slot in the event " $B$ and $E$ ."

Even-numbered slots that are black = $2, 4, 6, 8, 10, 24, 22, 20, 26, 28$

鈥� $鈥� P (B 鈭� E) = \frac{10}{38} = 0.2632$

Part (d) Step 1 : Given Information

We have to determine $P (B o r E)$

Part (d) Step 2 : Simplification

The chance of a ball falling on the black slot or an even numbered slot is denoted by the letter $P (B o r E)$ .

Because

P (B a n d E)

is greater than zero, event

B

and event

E

are not mutually exclusive.

When one adds

P (B)

and

P (E)

, the

P (B a n d E)

are included twice.

As a result,

P (B o r E) 鈮� P (B) + P (E)

$F r o m t h e a b o v e s u b p a r t s, P (B) = 0.4737, P (E) = 0.5, P (B a n d E) = 0.2632 . 鈥� 鈥� P (B o r E) = 0.4737 + 0.5 鈥� 0.2632 鈥� 鈥� P (B o r E) = 0.7105$