Respuesta :
Answer:
a) For this case we can find the cumulative distribution function first:
[tex] F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k[/tex]
So then by the complement rule we have this:
[tex]P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a[/tex]
b) [tex] P(Y>a)= q^a[/tex]
[tex] P(Y>b) = q^b[/tex]
So then we have this using independence:
[tex] P(Y> a+b) = q^{a+b}[/tex]
We want to find the following probability:
[tex] P(Y> a+b |Y>a) [/tex]
Using the definition of conditional probability we got:
[tex] P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b) [/tex]
And we see that if a = 2 and b=5 we have:
[tex] P(Y> 2+5 | Y>2) = P(Y>5)[/tex]
c) For this case we use independent identical and with the same distribution experiments.
And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.
Step-by-step explanation:
Previous concepts
The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"
[tex]P(X=x)=(1-p)^{x-1} p[/tex]
If we define the random of variable Y we know that:
[tex]Y\sim Geo (1-p)[/tex]
Part a
For this case we can find the cumulative distribution function first:
[tex] F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k[/tex]
So then by the complement rule we have this:
[tex]P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a[/tex]
Part b
For this case we can use the result from part a to conclude that:
[tex] P(Y>a)= q^a[/tex]
[tex] P(Y>b) = q^b[/tex]
So then we have this assuming independence:
[tex] P(Y> a+b) = q^{a+b}[/tex]
We want to find the following probability:
[tex] P(Y> a+b |Y>a) [/tex]
Using the definition of conditional probability we got:
[tex] P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b) [/tex]
And we see that if a = 2 and b=5 we have:
[tex] P(Y> 2+5 | Y>2) = P(Y>5)[/tex]
Part c
For this case we use independent identical and with the same distribution experiments.
And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.
The probabilities are as follows;
A. [tex]\rm F(k) = q^a[/tex]
B. [tex]\rm P(Y > 2+5|Y > 2) = P(Y > 5)[/tex]
C. For this case we use independent identical and with the same distribution experiment.
What is normal a distribution?
It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.
Let Y denote a geometric random variable with a probability of success p.
A. Show that for a positive integer a, P(Y > a) = qa.
[tex]\rm F(k) = P(Y\leq k) = \Sigma_{k'=1}^k P(Y=k') \\\\F(k) = \Sigma _{k'=1}^k p(1-p)^{k'-1} = \Sigma _{k'=1}^k 1-(1-p)^{k}[/tex]
By complementary rule, we have
[tex]\rm P(Y > a) = 1 - F(a) = 1-[1-(1-p)^a] = (1-p)^a = q^a[/tex]
B. Show that for positive integers a and b, P(Y > a+b|Y>a) = qb = P(Y>b).
[tex]\rm P(Y > a) = q^a\\\\P(Y > b) = q^b[/tex]
So then we have
[tex]P(Y > a+b) = q^{a+b}[/tex]
Then the probability will be
[tex]P(Y > a+b|Y > a) = \dfrac{P(Y > a +b\cap Y > a)}{P(Y > a)} \\\\P(Y > a+b|Y > a)= \dfrac{P(Y > a+b)}{P(Y > a)} = \dfrac{q^{a+b}}{q^a} = q^b = P(Y > b)[/tex]
And we see that if a = 2 and b = 5, we have
[tex]P(Y > 2+5|Y > 2) = P(Y > 5)[/tex]
C. For this case we use independent identical and with the same distribution experiment.
More about the normal distribution link is given below.
https://brainly.com/question/12421652
