Probability of X: Geo Distribution with 2/3

[songoku]

Homework Statement

A fair cubical die has two yellow faces and four blue faces. In one game, the die is rolled repeatedly until yellow face appears uppermost. The maximum number of throw is 4. The random variable X represents the number of times blue face appears uppermost and random variable Y represents the number of times the die is rolled.
a. Find P(X = 3)
b. Find the probability distribution of X
c. Find E(X)
d. Find P(Y = 4)
e. P(Y = X)
f. If the game is repeated 200 times, write down (do not evaluate) an expression for the probability obtaining X = 3 exactly 25 times.
g. Use appropriate approximation, calculate the probability of question (f)

Relevant Equations

Not sure

a. P(X = 3) = (4/6) x (4/6) x (4/6) x (2/6) = 8/81

b. X ~ Geo (2/3)
Is this correct?

Thanks

 

[PeroK]

Homework Statement:: A fair cubical die has two yellow faces and four blue faces. In one game, the die is rolled repeatedly until yellow face appears uppermost. The maximum number of throw is 4. The random variable X represents the number of times blue face appears uppermost and random variable Y represents the number of times the die is rolled.
a. Find P(X = 3)
b. Find the probability distribution of X
c. Find E(X)
d. Find P(Y = 4)
e. P(Y = X)
f. If the game is repeated 200 times, write down (do not evaluate) an expression for the probability obtaining X = 3 exactly 25 times.
g. Use appropriate approximation, calculate the probability of question (f)
Homework Equations:: Not sure

a. P(X = 3) = (4/6) x (4/6) x (4/6) x (2/6) = 8/81

b. X ~ Geo (2/3)
Is this correct?

Thanks

Yes, this is a set of Bernoulli trials with characteristic probability $\frac 2 3$:

https://en.wikipedia.org/wiki/Geometric_distribution

Note the difference and similarity between the distributions of $X$ and $Y$.

Sorry, I just saw that there is a maximum number of throws of $4$. It's a truncated geometric distribution.

 

[songoku]

Ok let me continue:

c. E(X) = q/p = 1/2

d. P( Y = 4) = (2/3)⁴ + (2/3)³ x (1/3) = 8/27

e. P (Y = X) = P(X = 4) = (2/3)⁴ = 16/81

f. Let random variable A represents number of games where X = 3
A ~ Binomial (n, p) so A ~ Binomial (200 , 8/81)
P (X = 25) = ²⁰⁰C₂₅ (8/81)²⁵ (73/81)⁷⁵

Are all correct up until this? Thanks

 

[PeroK]

Ok let me continue:

c. E(X) = q/p = 1/2

d. P( Y = 4) = (2/3)⁴ + (2/3)³ x (1/3) = 8/27

e. P (Y = X) = P(X = 4) = (2/3)⁴ = 16/81

f. Let random variable A represents number of games where X = 3
A ~ Binomial (n, p) so A ~ Binomial (200 , 8/81)
P (X = 25) = ²⁰⁰C₂₅ (8/81)²⁵ (73/81)⁷⁵

Are all correct up until this? Thanks

Note that there is a limit on the number of throws. I missed that. You need to calculate $E(X)$ for this.

When they ask for the distribution, I think they want the values for each outcome. There's a limit of four throws.

 

[songoku]

Note that there is a limit on the number of throws. I missed that. You need to calculate $E(X)$ for this.

When they ask for the distribution, I think they want the values for each outcome. There's a limit of four throws.

Not sure I understand what you mean but I did (c) using another way. I draw probability distribution table for X and I get this:
P(X = 0) = 1/3
P(X = 1) = 2/9
P(X = 2) = 4/27
P(X = 3) = 8/81
P(X = 4) = 16/81

Then E(X) = 0 . (1/3) + (1) (2/9) + (2) (4/27) + (3) (8/81) + (4) (16/81) = 130/81

Is this what you mean? I think this is the correct way to do (c). Thanks

 

[PeroK]

Not sure I understand what you mean but I did (c) using another way. I draw probability distribution table for X and I get this:
P(X = 0) = 1/3
P(X = 1) = 2/9
P(X = 2) = 4/27
P(X = 3) = 8/81
P(X = 4) = 16/81

Then E(X) = 0 . (1/3) + (1) (2/9) + (2) (4/27) + (3) (8/81) + (4) (16/81) = 130/81

Is this what you mean? I think this is the correct way to do (c). Thanks

Yes, that all looks good.

 

[songoku]

Yes, that all looks good.

How about (d), (e), and (f)?

 

[PeroK]

How about (d), (e), and (f)?

Yes, they look correct.

 

[songoku]

I think I can do (g)

Thank you very much perok

 

[archaic]

I tell you, the probability of death is 100%