Chapter 5.2, Problem 45E

(a)

To determine

To construct: A two way table displaying the sample space of events B and E.

Answer to Problem 45E

Event	Sample space
B	2, 35, 4, 33, 6, 31, 8, 29, 10, 13, 24, 15, 22, 17, 20, 11, 26, 28
E	0, 2, 4, 16, 6, 18, 8, 12, 10, 00, 36, 24, 34, 22, 32, 20, 30, 26, 28

Explanation of Solution

Given information:

Total number of slots in a roulette = 38

Number of red slots = 18

Number of black slots = 18

Number of green slots = 2

B is the event of a ball landing in a black slot

E is the event of a ball landing in an even numbered slot

Calculation:

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)

To determine

To compute: The probability of event B and event E.

Answer to Problem 45E

  $\text{P }\left(\text{B}\right)=0.\text{4737}$

  $\text{P }\left(\text{E}\right)=0.\text{5}$

Explanation of Solution

Formula used:

  $\text{P(B) = }\frac{\text{Total number of outcomes in the sample space in the event B}}{\text{Total number of outcomes}}$

  $\text{P(E) = }\frac{\text{Total number of outcomes in the sample space in the event E}}{\text{Total number of outcomes}}$

Calculation:

  $\text{P(B)}=\frac{18}{38}=0.4737$

  $\text{P(E)}=\frac{19}{38}=0.5$

(c)

To determine

To compute: The probability of the event “B and E”

Answer to Problem 45E

The probability of the event “B and E” is 0.2632.

Explanation of Solution

Formula used:

  $\text{P(B}\cap \text{E) = }\frac{\text{Total number of outcomes in the sample space in the event (B}\cap \text{E)}}{\text{Total number of outcomes}}$

Event “B and E” means the ball lands in the black even numbered slot.

Calculation:

Slots which are black even numbered $=\text{ 2},\text{ 4},\text{ 6},\text{ 8},\text{ 1}0,\text{ 24},\text{ 22},\text{ 2}0,\text{ 26},\text{ 28}$

  $\text{P}\left(\text{B}\cap \text{E}\right)=\frac{10}{38}=0.2632$

(d)

To determine

To compute: P (B or E)

Answer to Problem 45E

  $\text{P }\left(\text{B or E}\right)=0.\text{71}0\text{5}$

Explanation of Solution

Formula used:

  $\text{P }\left(\text{B or E}\right)=\text{ P }\left(\text{B}\right)+\text{ P }\left(\text{E}\right)–\text{ P }\left(\text{B and E}\right)$

P (B or E) means the probability of a ball landing on the black slot or on an even numbered.

As P (B and E) is not equal to zero, this means that event B and event E is not mutually exclusive.

By adding P (B) and P (E), one is including the P (B and E) twice.

Hence, P (B or E) $\ne$ P (B) + P (E)

Calculation:

From the above sub parts, $\text{P }\left(\text{B}\right)=0.\text{4737},\text{ P }\left(\text{E}\right)=0.\text{5},\text{ P }\left(\text{B and E}\right)=0.\text{2632}.$

  $\text{P }\left(\text{B or E}\right)=0.\text{4737 }+0.\text{5 }–0.\text{2632}$

  $\text{P }\left(\text{B or E}\right)=0.\text{71}0\text{5}$