Roulette An American roulette wheel has 38 slots with numbers 1through36,0,and 00, as shown in the figure. Of the numbered slots, 18are red, 18are black, and 2—the 0and 00—are 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. (Treat0and 00as even numbers.)

a. Make a two-way table that displays the sample space in terms of events Band E.

b. Find P(B)andP(E).

c. Describe the event “Band E” in words. Then find the probability of this event.

d. Explain why P(BorE)P(B)+P(E). Then use the general addition rule to compute P(BorE).

Short Answer

Expert verified

a. A two way table displaying the sample space of events Band 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= 0.4737and event E=0.5

c. The probability of the event “Band E” = 0.2632

d. The probability of the event “Bor E” =0.7105

Step by step solution

01

Part (a) Step 1 : Given Information

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

02

Part (a) Step 2 : Simplification

A roulette wheel has a total of 38slots.
There are 18red slots in all.
There are 18black slots in all.
The number of green slots is two.
The event Bis when a ball lands in a black slot.
The event Eis 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

03

Part (b) Step 1 : Given Information

We have to determine the probability of event Band event E.

04

Part (b) Step 2 : Simplification

We fill use the following formula :-

P(B)=TotalnumberofoutcomesinthesamplespaceintheeventBTotalnumberofoutcomesP(E)=TotalnumberofoutcomesinthesamplespaceintheeventETotalnumberofoutcomes

P(B)=1838=0.4737P(E)=1938=0.5

05

Part (c) Step 1 : Given Information

We have to determine the probability of the event “B and E

06

Part (c) Step 2 : Simplification

We will use the following formula :-

P(BE)=Totalnumberofoutcomesinthesamplespaceintheevent(BE)Totalnumberofoutcomes

The ball lands in the black even-numbered slot in the event "Band E."

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

P(BE)=1038=0.2632

07

Part (d) Step 1 : Given Information

We have to determineP(BorE)

08

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(BorE).

BecauseP(BandE)is greater than zero, event Band event Eare not mutually exclusive.
When one adds P(B)and P(E), theP(BandE)are included twice.
As a result, P(BorE)P(B)+P(E)

Fromtheabovesubparts,P(B)=0.4737,P(E)=0.5,P(BandE)=0.2632.P(BorE)=0.4737+0.50.2632P(BorE)=0.7105

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