# Thread: Probability and geometric distributions

1. ## Probability and geometric distributions

Calculate the following probabilities. I've been told that X~geom(0.1) means X follows a geometric distribution

(a) For X ~ Geom(0.1), calculate P(X ≤ 10).
(b) For X ~ Geom(0.2), calculate P(5 ≤ X ≤ 10).
(c) For X ~ P(5), calculate P(X < 6).

I know what the geometric formula is for a sequence, does this apply to this though? :S

2. Originally Posted by mitch_nufc
Calculate the following probabilities. I've been told that X~geom(0.1) means X follows a geometric distribution

(a) For X ~ Geom(0.1), calculate P(X ≤ 10).
(b) For X ~ Geom(0.2), calculate P(5 ≤ X ≤ 10).
(c) For X ~ P(5), calculate P(X < 6).

I know what the geometric formula is for a sequence, does this apply to this though? :S
Questions (a) and (b) require the formula for cumulative probability for a geometric distribution. See Geometric distribution - Wikipedia, the free encyclopedia:

$\Pr(X \leq k) = 1 - (1 - p)^k$.

(a) Substitute p = 0.1 and k = 10.

(b) $= \Pr(X \leq 10) - \Pr(X \leq 5)$. Calculate each of these just like (a) but use p = 0.2.

(c) Poisson distribution with a mean of 5: $\lambda = 5$. You need to substitute into the probability mass function (pmf). See Poisson distribution - Wikipedia, the free encyclopedia:

$\Pr(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.

Substitute $\lambda = 5$ and k = 0. Then do the same for k = 1, 2, 3, 4, 5. Then add up all these probabilities to get the answer.

3. (a) P(X<=10) = 1 - (0.9)^11 = 0.6862
(b) P(5<=X<=10) = 0.686 - P(X<=4) - 0.6862 - 0.4095 = 0.277
(c) P(X=k) = e^-5 . 5^k / k!
So P(X<6) = Probability(X=0 or 1 or ... or 5)
= e^5 (5/1! + 5^2/2! + ... + 5^5/5!) = 0.609

this is what ive got, do you agree?

4. Quote:
Originally Posted by mitch_nufc
Calculate the following probabilities. I've been told that X~geom(0.1) means X follows a geometric distribution

(a) For X ~ Geom(0.1), calculate P(X ≤ 10).
(b) For X ~ Geom(0.2), calculate P(5 ≤ X ≤ 10).
(c) For X ~ P(5), calculate P(X < 6).

I know what the geometric formula is for a sequence, does this apply to this though? :S

Questions (a) and (b) require the formula for cumulative probability for a geometric distribution. See Geometric distribution - Wikipedia, the free encyclopedia:

.

(a) Substitute p = 0.1 and k = 10.

(b) . Calculate each of these just like (a) but use p = 0.2.

(c) Poisson distribution with a mean of 5: . You need to substitute into the probability mass function (pmf). See Poisson distribution - Wikipedia, the free encyclopedia:

.

Substitute and k = 0. Then do the same for k = 1, 2, 3, 4, 5. Then add up all these probabilities to get the answer. __________________

shouldnt (b) be Pr(x < or equal to 10)-Pr(x< or equal to 4) as 5 should be inclusive in the answer. with mr fantastics theory we get 0.2203, but i think the answer should be 0.3022?

5. ## Geometric and binomial

a) For X~geom(0.1), calculate P(X≤10)
I got 0.6862 (someone told me this was wrong)

b)For X~geom(0.2), calculate P(5≤X≤10)
I got 0.2767 (again, someone told me im wrong)

c)For X~P(5), calculate P(X<6)
I got 0.6092 (again, someone told me i was wrong again!)

If you agree with mine then cool, if not can you show me how you got your answer?

6. Originally Posted by mr fantastic
Questions (a) and (b) require the formula for cumulative probability for a geometric distribution. See Geometric distribution - Wikipedia, the free encyclopedia:

$\Pr(X \leq k) = 1 - (1 - p)^k$.

(a) Substitute p = 0.1 and k = 10.

(b) $= \Pr(X \leq 10) - \Pr(X \leq {\color{red}4})$. Calculate each of these just like (a) but use p = 0.2.

(c) Poisson distribution with a mean of 5: $\lambda = 5$. You need to substitute into the probability mass function (pmf). See Poisson distribution - Wikipedia, the free encyclopedia:

$\Pr(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.

Substitute $\lambda = 5$ and k = 0. Then do the same for k = 1, 2, 3, 4, 5. Then add up all these probabilities to get the answer.
Correction of minor error/typo in red.

For (a) and (b), note that I have used the cumulative probability for a geometric distribution whose support does not include zero. If zero is included, you have to use the other cumulative probability.

Originally Posted by mitch_nufc

a) For X~geom(0.1), calculate P(X≤10)
I got 0.6862 (someone told me this was wrong)

b)For X~geom(0.2), calculate P(5≤X≤10)
I got 0.2767 (again, someone told me im wrong)

c)For X~P(5), calculate P(X<6)
I got 0.6092 (again, someone told me i was wrong again!)

If you agree with mine then cool, if not can you show me how you got your answer?
c) I get 0.61596. You probably need to re-check the calculations I said to do in my first post.

a) I get 0.65132 (assuming that X = 0 is not included in the support).

b) I differ with you here too.

Your troubles probably lie in the arithmetic, not the method.