Question

Consider the numberless roulette game at a casino. On a spin of the wheel, the ball lands in a space with color red (r), green (g), or black (b). The wheel has 19 red spaces, 19 green spaces, and 2 black spaces.
(a) In 40 spins of the wheel, find the probability of event A = {19 reds, 19 greens, and 2 blacks}.
(b) In 40 spins of the wheel, find the probability of the event G19 = {19 greens}.
(c) Given that you randomly choose to bet red and green only, what is the probability p that you bet a winner?

Answer 1

Answer:

A)0.0357

B) 0.1255

C) 0.475

Step-by-step explanation:

A) To find the probability of event A = {19 reds, 19 greens, and 2 blacks} in 40 spins, the formula is given as

P(R_19,G_19,B_2) = S•P(R)^(nr) •P(G)^(ng)•P(B)^(nb)

Where;

S = total number of outcomes

P(R) = Probability of a red space

P(G) = Probability of a green space

P(B) = Probability of a black space

nr = number of red spaces =19

ng = number of green spaces = 19

nb = number of black spaces = 2

So, let's calculate them;

S = 40!/(19!•19!•2!) = 27569305764000

P(R) = 19/40 ; P(G) = 19/40 ; P(B) = 2/40

Thus,P(R_19,G_19,B_2) = 27569305764000•(19/40)^(19)•(19/40)^(19)•(2/40)^(2) = 0.0357

B) Formula for the probability of the event G19 is given as;

P(G_19) = C(40,19)•P(G)^(ng)•P(O)^(no)

Where, P(O) is probability of other ball = 21/40

n(o) is number of other balls = 21

C(40,19) is 40 combination 19 using combination formula. Thus, C(40,19) = 40!/((19!)(21!)) = 131282408400

Thus,

P(G_19) = 131282408400•(19/40)^(19)•(21/40)^(21) = 0.1255

C)To solve this;

If we bet for red, we know that P(Red) = 19/40

Also, if we bet for green, we know that P(G) = 19/40

Thus probability of winning bet is;

P(winning bet) = (1/2)P(R) + (1/2)P(G)

P(winning bet) = (1/2)(19/40) + (1/2 (19/40) = 19/80 + 19/80 = 38/80 = 19/40 = 0.475

Answer 2

Answer:

a. Probability of Event A = 2.724037476607E−24

b. G19 = 0.123

c. P(Winning) = 0.475

Step-by-step explanation:

Given

Red (r) = 19

Green (g) = 19

Black (b) = 2

Total = 19 + 19 + 2 = 40

a. In 40 spins of the wheel, find the probability of event A = {19 reds, 19 greens, and 2 blacks}.

This is calculated as follows;

Let P(R) = Probability of Red

P(R) = 19/40

Let P(G) = Probability of Green

P(G) = 19/40

Let P(B) = Probability of Black

P(B) = 2/40

Total number of arrangement = 40!/(19!19!2!) = 27,569,305,764,000

Probability of Event A = 27,569,305,764,000 * (19/40)^19 * (19/40)^19 * (2/40)^19

Probability of Event A = 2.724037476607E−24

b. In 40 spins of the wheel, find the probability of the event G19 = {19 greens}.

Let P(G) = Probability of Green

P(G) = 19/40

Let P(Other) = Probability of any colour other than green = (2+19)/40

P(Other) = 21/40

Total = 40C19

G19 = 40C19 * (19/40)^19 * (21/40)^21

G19 = 0.125525075056335

G19 = 0.123 ---- Approximated

c. Given that you randomly choose to bet red and green only, what is the probability p that you bet a winner?

Let P(R) = Probability of Betting Red

P(R) = 19/40

Let P(G) = Probability of Betting Green

P(G) = 19/40

Let P(Winning) = Probability of Winning

P(Winning) = ½ * P(G) + ½ * P(R)

P(Winning) = ½ * 19/40 + ½ * 19/40

P(Winning) = ½(19/40 + 19/40)

P(Winning) = ½(38/40)

P(Winning) = 19/40

P(Winning) = 0.475