Question

In a video poker card game, the probability of a “winning” hand is 0.132. Determine the following (answer to 4 decimal places)

In 12 trials, what is the probability that you win exactly twice?

What is the probability that your first win is on your 6th hand?

What is the probability that you will win 2 or less of 12 hands?

What is the probability that you will win more than 3 of 12 hands?

How many times do you expect to play before your first winning hand?

In 25 games, how many do you expect to win?

What is the probability that you will win back to back hands?
And what assumption do you have to make in order to make this

calculation?

Answer #1

here this is binomial with parameter n=12 and
p=0.132 |

a)

probability that you win exactly twice
=P(X=2)=C(12,2)*(0.132)^{2}*(1-0.132)^{10}
=0.2792

b)

probability that your first win is on your 6th hand =P(lose 1st five games and wins 6th game)

=(1-0.132)^5*0.132 =0.0650

c)

probability that you will win 2 or less of 12 hands =P(X<2)=P(X=0)+P(X=1)

=C(12,0)*(0.132)^{0}*(1-0.132)^{12}
+C(12,1)*(0.132)^{1}*(1-0.132)^{11} =0.5167

d)

probability that you will win more than 3 of 12 hands :

P(X>3)=1-P(X<=3)= |
1-∑_{x=0}^{3}
(_{12}C_{x})0.132^{x}(1-0.132)^{(12-x)}
= |
0.0626 |

e)

expect to play before your first winning hand =1/p-1=1/0.132-1 =6.576

f)

In 25 games, how many do you expect to win =np=25*0.132 =3.3

g)

we assume that winning is independent from trial to trial.

probability that you will win back to back hands =0.132*0.132=0.0174

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