Probability of X: Geo Distribution with 2/3

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Homework Help Overview

The discussion revolves around a probability problem involving a fair cubical die with two yellow faces and four blue faces. The problem involves calculating probabilities related to the number of times blue faces appear when rolling the die until a yellow face appears, with a maximum of four rolls. The random variables X and Y represent the number of blue faces and the total number of rolls, respectively.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of probabilities for specific outcomes of X and Y, including the probability distribution and expected value. There is discussion about the nature of the distribution, with some noting it as a truncated geometric distribution due to the maximum number of throws.

Discussion Status

Several participants have provided calculations for probabilities and have engaged in verifying each other's work. There is an ongoing exploration of the expected value and distribution of X, with some participants suggesting alternative methods for calculating these values. The discussion is active, with participants questioning assumptions and clarifying the implications of the maximum number of throws.

Contextual Notes

There is a limit on the number of throws (maximum of 4), which influences the calculations and interpretations of the probability distributions. Participants are also navigating the distinction between geometric and truncated geometric distributions in their analyses.

songoku
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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
 
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songoku said:
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.
 
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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
 
songoku said:
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.
 
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PeroK said:
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
 
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songoku said:
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.
 
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PeroK said:
Yes, that all looks good.
How about (d), (e), and (f)?
 
songoku said:
How about (d), (e), and (f)?

Yes, they look correct.
 
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I think I can do (g)

Thank you very much perok
 
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I tell you, the probability of death is 100% :nb)
 

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