Probability Theory ; Binomial Distribution?

In summary, the problem involves flipping a coin until it comes up heads and rolling a die until it comes up with a 3 or 5. The number of flips needed is represented by X and the number of rolls needed is represented by Y. These values are assumed to be independent with both X and Y being greater than or equal to 1. The task is to determine the probability of X and Y being equal to a certain value, the expected value of X and Y, and the common value when X and Y are equal. The final expressions for the probability and expected value calculations are given.
  • #1
aeubz
16
0

Homework Statement


Now you and your fiend play a different game. You flip your coin until it comes up heads the first time. Let X denote the number of flips needed. Your friend rolls its die until it comes up "3" or "5". The first try let Y denote the number of rolls needed. Assume X and Y are independent; note that X >= 1 and Y >=1. a) Determine P(X=n), n>=1 b) Determine P(Y=n), n>=1 c) Determine E(x) d) Determine E(y) e) Determine P(x=y) f) Given that x = y, determine the expected value of this common value.

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Homework Equations


Binomial Distribution
nCk (P)^k (1-P)^n-k

The Attempt at a Solution


a) nC1 (1/2)(1/2)^n-1 (My friend did not include the nC1)
b) nC1 (1/3)(2/3)^n-1 (My friend did not include the nC1)
c) n*1/2
d) n*1/3
e) nC1 (1/2)(1/2)^n-1 * nC1 (1/3)(2/3)^n-1 (This is weird, my friend integrated from 1 to infinite) he got 1/4??
f) nsubx*psubx + nsuby*psuby (I definitely need help on this one.)
 
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  • #2
For a) and b), why do you think the nC1 should be in the expression? And for c) and d) the expectation value is a sum over all n. The final expression shouldn't have an 'n' in it. Maybe you'd better review the definition of 'expectation value'.
 
  • #3
Dick said:
For a) and b), why do you think the nC1 should be in the expression? And for c) and d) the expectation value is a sum over all n. The final expression shouldn't have an 'n' in it. Maybe you'd better review the definition of 'expectation value'.

Ok so the reason nC1 should not be there is because we only care about the n trial. Right? Still a bit confused about the Expected.. If X is geometric with probability p then E(X) = 1/p = 2 and E(Y) = 3? And e) is P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4

f) is still bothering me a little..
 
Last edited:
  • #4
Right. The first n-1 trials have to be failures, only the last one needs to succeed. And, yes, now I think you have the expectation values right as well. I've got to confess, I'm a little vague on the last one. Probabilility isn't my field, but isn't it sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1)?
 
  • #5
Dick said:
Right. The first n-1 trials have to be failures, only the last one needs to succeed. And, yes, now I think you have the expectation values right as well. I've got to confess, I'm a little vague on the last one. Probabilility isn't my field, but isn't it sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1)?

Ok ok, so by definition, E[X] = sum(n=1 to +oo) n * f(n) where f(n) is the density function. Yes I agree. Thank you very very much for all your help!
 

Related to Probability Theory ; Binomial Distribution?

What is probability theory?

Probability theory is a branch of mathematics that deals with the study of random events and the likelihood of their occurrence. It provides a framework for understanding and predicting the outcomes of uncertain situations.

What is a binomial distribution?

A binomial distribution is a probability distribution that describes the number of successes in a sequence of independent trials, where each trial has a fixed probability of success. It is often used to model the number of successes in a sample of size n from a larger population.

What are the assumptions of the binomial distribution?

The assumptions of the binomial distribution include: the trials are independent, each trial has only two possible outcomes (success or failure), the probability of success remains constant for each trial, and the sample size is fixed.

How is the binomial distribution calculated?

The binomial distribution can be calculated using the formula P(x) = (n choose x) * p^x * (1-p)^(n-x), where n is the sample size, x is the number of successes, and p is the probability of success for each trial. The "n choose x" term represents the number of ways to choose x successes from n trials.

What is the relationship between the binomial distribution and the normal distribution?

As the sample size increases, the binomial distribution approaches a normal distribution. This is known as the Central Limit Theorem. This relationship is useful because the normal distribution is easier to work with mathematically and can be used to approximate probabilities for the binomial distribution.

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